Sample Exercises

Hint

No, the gravitational force is a long-range force and we cannot escape from it.

Hint

Due to the small spaceship, the force of attraction between the earth and the ship is too less, the astronaut cannot detect gravity. If the size of the space station is very large, the magnitude of the gravity will also become appreciable and hence he can hope to detect it.

Hint

Hint

a. Decreases.
The value of acceleration due to gravity at a height ‘h’ above the ground ‘gh’ is given by, \(g_h=g\left[1-2 h / R_e\right]\) where \(R_e\) is radius of Earth. Therefore \(g_h\) decreases with an increase in height.
b. Acceleration decreases according to the formula \(\mathrm{g}^{\prime}=\mathrm{g}\left(1-\frac{\mathrm{d}}{\mathrm{R_e}}\right)\), where \(d\) is depth and \(R_e\) is the radius of the earth.
c. Mass of the body.
The acceleration due to gravity of any planet of mass ‘ \(M\) ‘ and radius ‘ \(R\) ‘ is given by, \(g\) \(=G M / R^2\), where, \(G\) is the universal gravitational constant. So, the value of ‘ \(g\) ‘ is dependent on the mass of the earth but independent of mass of the body.
d. More accurate.
The gravitational potential energy of any point at a distance ‘ \(r\) ‘ from the centre of the earth is \(V(r)=G m M / r\)
So, Potential Energy difference
\(
\begin{aligned}
&=V\left(r_2\right)-V\left(r_1\right) \\
&\Delta V=-G m M\left(\frac{1}{r_2}-\frac{1}{r_1}\right) \\
&=+G m M\left(\frac{\left(r_2-r_1\right)}{r_1 r_2}\right) \\
&=\left(\frac{G M}{r_1 r_2}\right) m\left(r_2-r_1\right) \\
&=\operatorname{gm}\left(r_2-r_1\right) \\
&=m g r_2-m g r_1
\end{aligned}
\)
(Assuming \(r_2-r_1 \approx R^2\) )
Thus \(-\mathrm{GmM}\left(1 / \mathrm{r}_2-1 / \mathrm{r}_1\right)\) is a more accurate relation.

Hint

Time taken by the Earth to complete one revolution around the Sun, \(\mathrm{T}_{\mathrm{e}}=1\) year
Time taken by the planet to complete one revolution around the Sun, \(T_P=\) \(\frac{1}{2} \mathrm{~T}_{\mathrm{e}}=\frac{1}{2}\) year
Orbital radius of the planet \(=\mathrm{Rp}\)
From Kepler’s third law of planetary motion, we can write:
\(
\left(\frac{\mathrm{R}_{\mathrm{P}}}{\mathrm{R}_{\mathrm{e}}}\right)^3=\left(\frac{\mathrm{T}_{\mathrm{P}}}{\mathrm{T}_{\mathrm{e}}}\right)^2
\)
\(
\left(\frac{R_P}{R_e}\right)=\left(\frac{T_P}{T_e}\right)^{2 / 3}
\)
\(
=0.5^{(2 / 3)}=0.63
\)
Hence, the orbital radius of the planet will be \(0.63\) times of the Earth.

Hint

Orbital period of \(I_0, T_{10}=1.769\) days \(=1.769 \times 24 \times 60 \times 60 \mathrm{~s}\)
Orbital radius of \(I_0, R_{10}=4.22 \times 10^8 \mathrm{~m}\)
Satellite \(I_0\) is revolving around the Jupiter
Mass of the latter is given by the relation:
\(
M_J=\frac{4 \pi^2 R_{\pm}^3}{G T_{10}^2} \ldots . \text { (i) }
\)
Where
\(M_J=\) Mass of Jupiter
\(G=\) Universal gravitational constant
Orbital period of the earth
\(T_e=365.25\) days \(=365.25 \times 24 \times 60 \times 60 \mathrm{~s}\)
Orbital radius of the Earth
\(R_e=1 A U=1.496 \times 10^{11} \mathrm{~m}\)
Mass of Sun is given as
\(
M_s=\frac{4 \pi^2 R_e^3}{G T_e^2} \text {.. (ii) }
\)
\(
\begin{aligned}
&\therefore \frac{M_s}{M_J}=\frac{4 \pi^2 R_e^3}{G T_e^2} \times \frac{G T_{I o}^2}{4 \pi^2 R_{\mathrm{lo}}^3}=\frac{R_e^3}{R_{\mathrm{lo}}^3} \times \frac{T_{\mathrm{lo}}^2}{T_e^2} \\
&=\left(\frac{1.769 \times 24 \times 60 \times 60}{365.25 \times 24 \times 60 \times 60}\right)^2 \times\left(\frac{1.496 \times 10^{11}}{4.22 \times 10^8}\right) \\
&=1045.04 \\
&\therefore \frac{M_s}{M_J} \sim 1000 \\
&M_s \sim 1000 \times M_J
\end{aligned}
\)
Hence, it can be inferred that the mass of Jupiter is about one thousandth that of the Sun.

Hint

Here, \(r=50000 \mathrm{ly}=50000 \times 9.46 \times 10^{15} \mathrm{~m}=4.73 \times 10^{20} \mathrm{~m}\) \(M=2.5 \times 10^{11}\) solar mass \(=2.5 \times 10^{11} \times\left(2 \times 10^{30}\right) \mathrm{kg}=5.0 \times 10^{41} \mathrm{~kg}\)
We know that
\(
\begin{aligned}
&M=\frac{4 \pi^2 r^3}{G T^2} \\
&\text { or } T=\left(\frac{4 \pi^2 r^3}{\mathrm{GM}}\right)^{\frac{1}{2}}=\left[\frac{4 \times\left(\frac{22}{7}\right)^2 \times\left(4.73 \times 10^{20}\right)^3}{6.67 \times 10^{-11} \times\left(5.0 \times 10^{41}\right)}\right]^{\frac{1}{2}} \\
&=1.12 \times 10^{16} \mathrm{~s}
\end{aligned}
\)

Hint

(a) We know,
Potential energy of satellite(P.E) \(=-G M m / r\)
Kinetic energy of satellite(K.E) \(=\mathrm{GMm} / 2 \mathrm{r}\)
Total energy \(=\) P.E \(+\) K.E
\(=-\mathrm{GMm} / \mathrm{r}+\mathrm{GMm} / 2 \mathrm{r}\)
\(=-G M m / 2 r=-K . E\)
e.g the total energy of an orbiting satellite is negative of the kinetic energy.

(b) the energy required to rocket an orbiting satellite out of earth’s Gravitational influence is less than the energy required to project a stationary object, because, in the case of orbiting satellite the Gravitational pull of the earth acting on it is balanced by centripetal force so, no work is required against the Gravitational pull.

Hint

Escape velocity is given by,
\(
\begin{aligned}
V_e &=\sqrt{2 g r}=\sqrt{\frac{2 G M}{r}} \\
&=\sqrt{\frac{2 G M}{R+h}}
\end{aligned}
\)
where,
\(\mathrm{g}\) is the acceleration due to gravity at the point from where the body is projected,
r is the distance from the centre of the earth from where the body is projected, and
\(h\) is the height from the surface of the earth.
(a) Escape velocity, \(v_e\) does not depend on the mass of the body.
(b) Escape velocity, \(v_e\) depends on upon \(g\) and at different locations \(g\) is different \(\left(g_\lambda=g\right.\)
\(R \omega^2 \cos ^2 \lambda\) ), therefore, \(v_e\) slightly depends on upon the angle of latitude \(\left(\because R \omega^2 \cos ^2 \lambda\right.\) \(\leq 3.36 \mathrm{~cm} / \mathrm{s}^2\) ).
(c) Escape velocity, \(v_e\) does not depend on upon the direction of projection.
(d) Escape velocity, \(\mathrm{v}_{\mathrm{e}}\) depends on upon height from the surface of the earth.

Hint

The torque exerted by the sun on the comet is zero.
Therefore, angular momentum of comet remains constant.
i.e., \(\quad \mathrm{mvr}=\mathrm{constant}\)
\(\Rightarrow \quad \mathrm{vr}=\) constant [because \(\mathrm{m}\) is constant].
In an elliptical orbit distance of the comet from the sun changes, therefore, the speed of the comet also changes.
(a) Linear speed of the comet is not constant.
(b) Angular speed \(\omega\) of the comet is not constant.
(c) Angular momentum of the comet is constant.
(d) Kinetic energy of comet \(1 / 2 \mathrm{mv}^2\) changes because the speed of the comet changes.
(e) Potential energy of the comet is \(-\frac{G M m}{r}\)

(f) Total energy of the comet remains constant.

Hint

(a) Legs hold the entire mass of a body in a standing position due to gravitational pull. In space, an astronaut feels weightlessness because of the absence of gravity. Therefore, swollen feet of an astronaut do not affect him/her in space.
(b) A swollen face is caused generally because of apparent weightlessness in space. Sense organs such as eyes, ears nose, and mouth constitute a person’s face. This symptom can affect an astronaut in space.
(c) Headaches are caused because of mental strain. It can affect the working of an astronaut in space.
(d) Space has different orientations. Therefore, the orientational problem can affect an astronaut in space.

Hint

We know that the gravitational potential \((\mathrm{V})\) in a sphere is constant throughout. Also, the gravitational intensity will act in the downward direction as its upper half is cut (gravitational force will also act in a downward direction). Hence the required direction of gravitational intensity is shown by arrow c.

Hint

Since, the gravitational intensity is defined as the negative of the gradient of gravitational potential i.e., \(\left(-\frac{d v}{d r}\right)\) and the value of gravitational potential inside the spherical shell is constant, the value of gravitational intensity is zero. Consider any point \(\mathrm{p}\) as shown, inside a sphere then the net gravitational intensity at that point is zero. If the upper half is cut off, the pull in the left and right side cancel each other just like the full sphere base but the pull upwards is missing. Thus the effective force is downwards along \(e\).

Hint

Mass of the Sun, \(M_s=2 \times 10^{30} \mathrm{~kg}\)
Mass of the Earth, \(\mathrm{M}_{\mathrm{e}}=6 \times 10^{24} \mathrm{~kg}\)
Orbital radius, \(\mathrm{r}=1.5 \times 10^{11} \mathrm{~m}\)
Mass of the rocket \(=\mathrm{m}\)
Let \(\mathrm{x}\) be the distance from the centre of the Earth where the gravitational force acting on satellite \(\mathrm{P}\) becomes zero.

From Newton’s law of gravitation, we can equate gravitational forces acting on satellite \(\mathrm{P}\) under the influence of the Sun and the Earth as:
\(
\frac{\mathrm{GmM}_s}{(\mathrm{r}-\mathrm{x})^2}=\frac{\mathrm{GmM}_e}{\mathrm{x}^2}
\)
\(
\left[\frac{(\mathrm{r}-\mathrm{x})}{\mathrm{x}}\right]^2=\mathrm{M}_s / \mathrm{M}_e
\)
\(
(r-x) / x=\left(\frac{2 \times 10^{30}}{60 \times 10^{24}}\right)^{1 / 2}=577.35
\)
\(
1.5 \times 10^{11}-\mathrm{x}=577.35 \mathrm{x}
\)
\(
578.35 \mathrm{x}=1.5 \times 10^{11}
\)
\(
x=1.5 \times 10^{11} / 578.35=2.59 \times 10^8 \mathrm{~m}
\)

Hint

Orbital radius of the Earth around the Sun, \(r=1.5 \times 10^{11} \mathrm{~m}\)
Time taken by the Earth to complete one revolution around the Sun,
\(
\begin{aligned}
&T=1 \text { year }=365.25 \text { days } \\
&=365.25 \times 24 \times 60 \times 60 s
\end{aligned}
\)
Universal gravitational constant, \(G=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\)
Thus, mass of the Sun can be calculated using the relation,
\(
\begin{aligned}
&M=\frac{4 \pi^2 r^3}{G T^2} \\
&=\frac{4 \times(3.14)^2 \times\left(1.5 \times 10^{11}\right)^3}{\left(6.67 \times 10^{11} \times(365.25 \times 24 \times 60 \times 60)\right)^2} \\
&=\frac{133.24 \times 10}{6.64 \times 10^4}=2.0 \times 10^{30} \mathrm{~kg}
\end{aligned}
\)
Hence, the mass of the Sun is \(2 \times 10^{30} \mathrm{~kg}\).

Hint

Distance of Earth from Sun, \(\mathrm{r}_{\mathrm{e}}=1.5 \times 10^{11} \mathrm{~m}\)
Time period of Earth \(=\mathrm{T}_{\mathrm{e}}\)
Time period of Saturn, \(T_s=29.5 T_e\)
Distance of Saturn from the Sun \(=r_{\mathrm{s}}\)
From Kepler’s third law of planetary motion,
\(
\begin{aligned}
&\mathrm{T}=\sqrt{\frac{4 \pi^2 \mathrm{r}^3}{\mathrm{GM}}} \\
&\Rightarrow \mathrm{r}_{\mathrm{s}}^3 / \mathrm{r}_{\mathrm{e}}^3=\mathrm{T}_{\mathrm{s}}^2 / \mathrm{T}_{\mathrm{e}}^2
\end{aligned}
\)
\(
r_s=r_e\left(T_s / T_e\right)^{2 / 3}
\)
\(
=1.5 \times 10^{11} \times 29.5^{2 / 3}
\)
\(
=14.32 \times 10^{11} \mathrm{~m}
\)
Therefore, the distance between Saturn and the Sun is \(1.43 \times 10^{12} \mathrm{~m}\).

Alternate:

By Keplers law: \(T^2 \propto r^3\)
\(
\begin{aligned}
& \frac{T_S^2}{T_E^2}=\frac{r_S^3}{r_E^3} \\
\therefore \quad r_s &=\left(\frac{T_S}{T_E}\right)^{2 / 3} \times r_E \\
&=\left(\frac{29.4 \times T_E}{T_E}\right)^{2 / 3} \times 1.5 \times 10^8 \mathrm{~km} \\
&=1.43 \times 10^9 \mathrm{~km}
\end{aligned}
\)

Hint

Weight of the body, \(\mathrm{W}=63 \mathrm{~N}\)
Acceleration due to gravity at height \(h\) from the Earth’s surface is given by the relation:
\(
\mathrm{g_h}=\frac{\mathrm{g}}{\left(1+\frac{\mathrm{h}}{\mathrm{R}_{\mathrm{e}}}\right)^2}
\)
Substituting \(\mathrm{h}=\mathrm{R}_{\mathrm{e}} / 2\)
\(
\mathrm{g_h}=4 \mathrm{~g} / 9
\)
Weight of body of mass \(m\) at height \(h\) is given as:
\(
\begin{aligned}
\mathrm{W_h} &=\mathrm{mg_h} \\
&=4 / 9 \mathrm{mg} \\
&=4 / 9 \times 63=28 \mathrm{~N}
\end{aligned}
\)

Hint

Weight of a body of mass \(m\) at the Earth’s surface, \(W=m g=250 \mathrm{~N}\)
Body of mass \(m\) is located at depth, \(\mathrm{d}=\frac{1}{2} R_e\)
Where
\(R_e=\) Radius of Earth
Acceleration due to gravity at depth \(g(d)\) is given by the relation:
\(
g(d)=\left(1-\frac{d}{R_e}\right) g
\)
\(
=\left(1-\frac{R_e}{2 \times R_e}\right) g=\frac{1}{2} g
\)
Weight of the body at depth \(d\),
\(
\begin{aligned}
&\mathrm{W(d)}={mg(d)} \\
&=m \times \frac{1}{2} g=\frac{1}{2} m g=\frac{1}{2} W \\
&=\frac{1}{2} \times 250=125 \mathrm{~N}
\end{aligned}
\)

Hint

Initial kinetic energy of rocket \(=1 / 2 \mathrm{mv}^2=1 / 2 \times \mathrm{m} \times(5000) 2=1.25 \times 10^7 \mathrm{~mJ}\)
At distance \(r\) from centre of earth, kinetic energy becomes zero
Change in kinetic energy \(=1.25 \times 10^7-0=1.25 \times 10^7 \mathrm{~m} \mathrm{~J}\)
This energy changes into potential energy.
Initial potential energy at the surface of earth \(=\mathrm{GM}_{\mathrm{e}} \mathrm{m} / \mathrm{r}\)
\(=\frac{-\left(6.67 \times 10^{-11}\right) \times\left(6 \times 10^{24}\right) m}{6.4 \times 10^6}=-6.25 \times 10^7 \mathrm{~J}\)
Final potential energy at distance \(r=-\frac{G M_e m}{r}\)
\(=\frac{-\left(6.67 \times 10^{-11}\right) \times\left(6 \times 10^{24}\right) m}{r}=-4 \times 10^{14} \frac{m}{r} J\)
\(\therefore\) Change in potential energy \(=6.25 \times 10^7 \mathrm{~m}-4 \times 10^{14} \frac{\mathrm{m}}{\mathrm{r}}\)
Using the law of conservation of energy
\(6.25 \times 10^7 \mathrm{~m}-\frac{4 \times 10^{14} \mathrm{~m}}{\mathrm{r}}=1.25 \times 10^7 \mathrm{~m}\)
i.e \(r=\frac{4 \times 10^{14}}{5 \times 10^7} m=8 \times 10^{16} \mathrm{~m}\)

Hint

Escape velocity of a projectile from the Earth, \(v_{\mathrm{esc}}=11.2 \mathrm{~km} / \mathrm{s}\)
Projection velocity of the projectile, \(v_{\mathrm{p}}=3 \mathrm{v}_{\mathrm{esc}}\)
Mass of the projectile \(=m\)
Velocity of the projectile far away from the Earth \(=v_{\mathrm{f}}\)
Total energy of the projectile on the Earth \(=\left(\frac{1}{2}\right) m v_{\mathrm{p}}^2-\left(\frac{1}{2}\right) m v_{\mathrm{esc}}{ }^2\)
Gravitational potential energy of the projectile far away from the Earth is zero.
Total energy of the projectile far away from the Earth \(=\left(\frac{1}{2}\right) m v^2\)
From the law of conservation of energy, we have
\(
\begin{aligned}
\left(\frac{1}{2}\right) m v_{\mathrm{p}}^2-\left(\frac{1}{2}\right) m v_{\mathrm{esc}}{ }^2=\left(\frac{1}{2}\right) m v_{\mathrm{f}}^2 \\
v_{\mathrm{f}} =\left(v_{\mathrm{p}}^2-v_{\mathrm{esc}}{ }^2\right)^{1 / 2} \\
=\left[\left(3 v_{\mathrm{esc}}\right)^2-v_{\mathrm{esc}}{ }^2\right]^{1 / 2} \\
=\sqrt{8} v_{\mathrm{esc}} \\
=\sqrt{8} \times 11.2 \\
=31.68 \mathrm{~km} / \mathrm{s} .
\end{aligned}
\)

Hint

Mass of the Earth, \(M=6.0 \times 10^{24} \mathrm{~kg}\)
Mass of the satellite, \(m=200 \mathrm{~kg}\)
Radius of the Earth, \(R_{\mathrm{e}}=6.4 \times 10^6 \mathrm{~m}\)
Universal gravitational constant, \(G=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\)
Height of the satellite, \(h=400 \mathrm{~km}=4 \times 10^5 \mathrm{~m}=0.4 \times 10^6 \mathrm{~m}\)
Total energy of the satellite at height \(h=\frac{1}{2} m v^2+\left(\frac{-\mathrm{G} M_e m}{R_{\mathrm{e}}+h}\right)\)
Orbital velocity of the satellite, \(v=\sqrt{\frac{\mathrm{G} M_e}{R_{\mathrm{e}}+h}}\)
Total energy of height, \(h=\frac{1}{2} m\left(\frac{\mathrm{G} M_{\mathrm{e}}}{R_{\mathrm{e}}+h}\right)-\frac{\mathrm{G} M_{\mathrm{e}} m}{R_{\mathrm{e}}+h}=-\frac{1}{2}\left(\frac{\mathrm{G} M_{\mathrm{e}} m}{R_{\mathrm{e}}+h}\right)\)
The negative sign indicates that the satellite is bound to the Earth. This is called bound energy of the satellite.

The energy required to send the satellite out of its orbit \(=-\) (Bound energy)
\(
\begin{aligned}
&=\frac{1}{2} \frac{\mathrm{G} M_{\mathrm{e}} m}{\left(R_{\mathrm{e}}+h\right)} \\
&=\frac{1}{2} \times \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 200}{\left(6.4 \times 10^6+0.4 \times 10^6\right)} \\
&=\frac{1}{2} \times \frac{6.67 \times 6 \times 2 \times 10}{6.8 \times 10^6}=5.9 \times 10^9 \mathrm{~J}
\end{aligned}
\)

Hint

Here, mass of each star, \(M=2 \times 10^{30} \mathrm{~kg}\)
Initial potential between two stars, \(r=10^9 \mathrm{~km}=10^{12} \mathrm{~m}\).
Initial potential energy of the system \(=-\mathrm{GMm} / \mathrm{r}\)
Total K.E. of the stars \(=1 / 2 \mathrm{Mv}^2+1 / 2 \mathrm{Mv}^2\)
where \(v\) is the speed of stars with which they collide. When the stars are about to collide, the distance between their centres, \(r^{\prime}=2 \mathrm{R}\).
:. Final potential energy of two stars \(=-\mathrm{GMm} / 2 \mathrm{R}\)
Since gain in K.E. is at the cost of loss in P.E
\(
\therefore M v^2=-\frac{G M M}{r}-\left(-\frac{G M M}{2 R}\right)=-\frac{G M M}{r}+\frac{G M M}{2 R}
\)
or \(2 \times 10^{30} v^2=-\frac{6.67 \times 10^{-11} \times\left(2 \times 10^{30}\right)^2}{10^{12}}+\frac{6.67 \times 10^{-11} \times\left(2 \times 10^{30}\right)^2}{2 \times 10^7}\)
\(
\begin{aligned}
&=-2.668 \times 10^{38}+1.334 \times 10^{43} \\
&=1.334 \times 10^{43} \mathrm{~J} \\
&\mathrm{v}=\frac{\sqrt{1.334 \times 10^{43}}}{2 \times 10^{30}}=2.58 \times 10^6 \mathrm{~ms}^{-1}
\end{aligned}
\)

Hint

Here \(G=6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2} ; M=100 \mathrm{~kg} ; R=0.1 \mathrm{~m}\), the distance between the two spheres, \(d=1.0 \mathrm{~m}\); Suppose that the distance of either sphere from the mid-point of the line joining their centre is \(r\). Then \(r=d / 2=0.5 \mathrm{~m}\). The gravitational field at the mid-point due to two spheres will be equal and opposite.
Hence the resultant gravitational field at the mid poiint \(=0\)
The gravitational potential at the mid point \(=\left(-\frac{\mathrm{GM}}{r}\right) \times 2\)
\(=-\frac{6.67 \times 10^{-11} \times 100 \times 2}{0.5}=-2.668 \times 10^{-8} \mathrm{Jkg}^{-1}\)
The object placed at that point would be in stable equilibrium. This is because any change in the position of the object will change the effective force in that direction.

Hint

Let \(\mathrm{R}\) be the radius of the orbit of Mars and \(\mathrm{R}^{\prime}\) be the radius of Mars. \(\mathrm{M}\) be the mass of the Sun and \(\mathrm{M}^{\prime}\) be the mass of Mars. If \(m\) is the mass of the space-ship, then the Potential energy of the space-ship due to the gravitational attraction of the Sun \(=-G M \mathrm{~m} / \mathrm{R}\) Potential energy of space-ship due to the gravitational attraction of Mars = \(G M^{\prime} m / R^{\prime}\) Since the K.E. of space ship is zero, therefore total energy of spaceship \(=-\frac{G M m}{R}-\frac{G M \prime m}{R}=-G m\left(\frac{M}{R}+\frac{M \prime}{R \prime}\right)\)
\(\therefore\) Energy required to rocket out the spaceship from the solar system = -(total energy of spaceship)
\(
\begin{aligned}
&=-\left[-G m\left(\frac{M}{R}+\frac{M \prime}{R} \prime\right)\right]=G m\left[\frac{M}{R}+\frac{M \prime}{R} \prime\right] \\
&=6.67 \times 10^{-11} \times 1000 \times\left[\frac{2 \times 10^{30}}{2.28 \times 10^{11}}+\frac{6.4 \times 10^{23}}{3395 \times 10^3}\right] \\
&=6.67 \times 10^{-8}\left[\frac{20}{2.28}+\frac{6.4}{33.95}\right] \times 10^{18} \mathrm{~J}=5.98 \times 10^{11} \mathrm{~J}
\end{aligned}
\)

Hint

Initial K.E \(=\frac{1}{2} m v^2 ;\) Initial P.E \(=-\frac{G M m}{R}\)
Where \(m=\) Mass of rocket, \(M=\) Mass of Mars, \(R=\) Radius of Mars.
\(\therefore\) Total initial energy \(=\frac{1}{2} m v^2-\frac{G M m}{R}\)
Since \(20 \%\) of \(k . E\) is lost,only \(80 \%\) remain to reach the height
\(\therefore\) Total inital energy available
\(
\begin{aligned}
&=\frac{4}{5} \times \frac{1}{2} m v^2-\frac{G M m}{R} \\
&=0.4 m v^2-\frac{G M m}{R}
\end{aligned}
\)
When the rocket reaches the highest point at a height \(h\) above the surface its K.E is zero and P.E \(=-\frac{G M m}{R+h}\)
Using the principle of conversion of energy
\(0.4 m v^2-\frac{G M m}{R}=\frac{G M m}{R+h}\)
or \(\frac{G M m}{R+h}=\frac{G M m}{R}-0.4 m v^2 \Rightarrow \frac{G M}{R+h}=\frac{G M}{R}-0.4 v^2\)
or \(\frac{G M}{R+h}=\frac{1}{R}\left[G M-0.4 R v^2\right] \Rightarrow \frac{R+h}{R}=\frac{G M}{G M-0.4 R v^2}\)
or \(\frac{h}{R}=\frac{G M}{G M-0.4 R v^2}-1\)
or \(\frac{h}{R}=\frac{0.4 R v^2}{G m-0.4 R v^2} \Rightarrow h=\frac{0.4 R^2 v^2}{G M-0.4 R v^2}\)
or \(h=\frac{0.4 \times\left(2 \times 10^3\right)^2 \times\left(3.395 \times 10^6\right)^2}{6.67 \times 10^{-11} \times 6.4 \times 10^{23}-0.4 \times\left(2 \times 10^3\right)^2 \times 3.395 \times 10^6} \mathrm{~m}\) \(=495 \mathrm{~km}\)

Hint

Mass of the Earth, \(M=6.0 \times 10^{24} \mathrm{~kg}\)
Radius of the Earth, \(R=6400 \mathrm{~km}=6.4 \times 10^6 \mathrm{~m}\)
Height of a geostationary satellite from the surface of the Earth, \(h=36000 \mathrm{~km}=3.6 \times 10^7 \mathrm{~m}\)

Gravitational potential energy due to Earth’s gravity at height \(h\),
\(
\begin{aligned}
&=\frac{-G M}{R+h} \\
&=\frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{3.6 \times 10^7+0.64 \times 10^7} \\
&=-\frac{6.67 \times 6}{4.24} \times 10^{13-7} \\
&=-9.4 \times 10^6 \frac{J}{K} g
\end{aligned}
\)

Hint

Yes
A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.
Gravitational force, \(f_g=\frac{G M m}{R^2}\)
Where
\(M=\) Mass of the star \(=2.5 \times 2 \times 10^{30}=5 \times 10^{30} \mathrm{~kg}\)
\(m=\) Mass of the body
\(R=\) Radius of the star \(=12 \mathrm{~km}=1.2 \times 10^4 \mathrm{~m}\)
\(\therefore f_g=\frac{6.67 \times 10^{-11} \times 5 \times 10^{30} \times m}{\left(1.2 \times 10^4\right)^2}=2.31 \times 10^{11} \mathrm{mN}\)
Centrifugal force, \(f_{\mathrm{c}}=m r \omega^2\)
\(\omega=\) Angular speed \(=2 \pi v\)
\(v=\) Angular frequency \(=1.2 \mathrm{rev} \mathrm{s}^{-1}\)
\(f_c=m R(2 \pi v)^2\)
\(=m \times\left(1.2 \times 10^4\right) \times 4 \times(3.14)^2 \times(1.2)^2=1.7 \times 10^5 \mathrm{mN}\)
Since \(f_g>f_{c^{\prime}}\) the body will remain stuck to the surface of the star

Hint

The gravitational force of attraction,
\(
\begin{aligned}
&F=\frac{G M m}{r^2} \\
&=\frac{6.67 \times 10^{-11} \times 10 \times 10}{(0.1)^2}=6.67 \times 10^{-7} \mathrm{~N}
\end{aligned}
\)

Hint

To calculate the gravitational force on ‘m’ due to other force,

Distance of the corner from the mid point of the square is half of the diagonal
So, \(r=\frac{a \sqrt{2}}{2}=\frac{a}{\sqrt{2}}\)
Now,Force due to the mass \(m\) placed at the corner \(A\)
\(\vec{F}_{O A}=\frac{G . m \cdot m}{r^2}\)
Force due to the mass \(2 \mathrm{~m}\) placed at the corner \(B\)
\(\vec{F}_{O B}=\frac{G \cdot m \cdot 2 m}{r^2}\)
Force due to the mass \(3 \mathrm{~m}\) placed at the corner \(\mathrm{C}\)
\(\vec{F}_{\text {OC }}=\frac{G . m .3 \mathrm{~m}}{\mathrm{r}^2}\)
Force due to the mass \(4 \mathrm{~m}\) placed at the corner \(D\) and \(\vec{F}_{O D}=\frac{G . m \cdot 4 m}{r^2}\)
Now,
Resultant force \(=\vec{F}_{O A}+\vec{F}_{O B}+\vec{F}_{O C}+\vec{F}_{O D}\)
Or,
We have already calculated the value of distance \(r=\frac{a \sqrt{2}}{2}=\frac{a}{\sqrt{2}}\)
\(
=\frac{2 G m m}{a^2}\left[-\frac{\vec{i}}{\sqrt{2}}+\frac{\vec{j}}{\sqrt{2}}\right]+\frac{4 G m m}{a^2}\left[\frac{\vec{i}}{\sqrt{2}}+\frac{\vec{j}}{\sqrt{2}}\right]+\frac{6 G m m}{a^2}\left[\frac{\vec{i}}{\sqrt{2}}-\frac{\vec{j}}{\sqrt{2}}\right]+\frac{8 G m m}{a^2}\left[\frac{-\vec{i}}{\sqrt{2}}-\frac{-\vec{j}}{\sqrt{2}}\right]
\)
After solving
\(F=\frac{4 \sqrt{2} G^2}{a^2}\).
Hence, the force at the mass placed in the middle of the square is \(F=\frac{4 \sqrt{2} \mathrm{Gm}^2}{a^2}\).

Hint

(a) Consider that mass ‘ \(m\) ‘ is placed at the midpoint \(O\) of side \(A B\) of equilateral triangle \(A B C\).
\(
\begin{aligned}
&\mathrm{AO}=\mathrm{BO}=\frac{a}{2} \\
&\text { Then } \vec{F}_{O A}=\frac{4 G m^2}{a^2} \text { along } \mathrm{OA}
\end{aligned}
\)
Also, \(\vec{F}_{O B}=\frac{4 G m^2}{a^2}\) along \(O B\)
\(
\begin{aligned}
&\mathrm{OC}=\frac{\sqrt{3} a}{2} \\
&\vec{F}_{O C}=\frac{4 G m^2}{\left\{(3) a^2\right\}}=\frac{4 G m^2}{3 a^2} \text { along OC }
\end{aligned}
\)
The net force on the particle at \(\mathrm{O}\) is \(\vec{F}=\vec{F}_{O A}+\vec{F}_{O B}+\vec{F}_{O C}\)
Since equal and opposite forces cancel each other, we have :
\(
\vec{F}=\vec{F}_{O C}=\frac{\mathrm{Gm}^2}{[(\sqrt{3} / 2) \mathrm{a}]^2}=\frac{4 \mathrm{Gm}^2}{3 \mathrm{a}^2} \text { along OC }
\)

(b) If the particle placed at \(O\) (centroid)
All the forces are equal in magnitude but their directions are different as shown in the figure. Equal and opposite forces along \(\mathrm{OM}\) and \(\mathrm{ON}\) cancel each other.
i.e., \(F \cos 30^{\circ}=F \cos 30^{\circ}\)
\(\therefore\) Resultant force \(=F-2 F \sin 30=0\)

Hint

Three spheres are placed with their centres at \(\mathrm{A}, \mathrm{B}\), and \(\mathrm{C}\) as shown in the figure.

Gravitational force on sphere \(C\) due to sphere \(B\) is given by
\(\overrightarrow{F_{C B}}=\frac{G m^2}{4 a^2} \cos 60^{\circ} \hat{i}+\frac{G m^2}{4 a^2} \cdot \sin 60^{\circ} \hat{j}\)
Gravitational force on sphere \(C\) due to sphere A is given by \(\vec{F}_{C A}=-\frac{G m^2}{4 a^2} \cos 60^{\circ} \hat{i}+\frac{G m^2}{4 a^2} \cdot \sin 60^{\circ} \hat{j}\)
\(\therefore \vec{F}_{C B}=\vec{F}_{C B}+\vec{F}_{C A}\)
\(
\begin{aligned}
&=+\frac{2 G m^2}{4 a^2} \sin 60^{\circ} \hat{j} \\
&=+\frac{2 G m^2}{4 a^2} \times \frac{\sqrt{3}}{2}
\end{aligned}
\)
i.e., magnitude \(=\frac{\sqrt{3} G m^2}{4 a^2}\) along \(C O\)

Hint

Force on \(M\) at \(C\) due to gravitational attraction.

\(
\begin{aligned}
&\overrightarrow{F_{C B}}=\frac{G m^2}{2 R^2} \hat{j} \\
&\overrightarrow{F_{C D}}=\frac{-G M^2}{4 R^2} \hat{i} \\
&\overrightarrow{F_{C A}}=\frac{-G M^2}{4 R^2} \cos 45 \hat{j}+\frac{G M^2}{4 R^2} \sin 45 \hat{j} \\
&\text { So, resultant force on } C, \\
&\therefore \overrightarrow{F_C}=\overrightarrow{F_{C A}}+\overrightarrow{F_{C B}}+\overrightarrow{F_{C D}} \\
&=-\frac{G M^2}{4 R^2}\left(2+\frac{1}{\sqrt{2}}\right) \hat{i}+\frac{G M^2}{4 R^2}\left(2+\frac{1}{\sqrt{2}}\right) \hat{j} \\
&\therefore F_C=\frac{G M^2}{4 R^2}(2 \sqrt{2}+1)
\end{aligned}
\)
For moving along the circle, \(\vec{F}=\frac{m v^2}{R}\)
\(
\text { or } \frac{G M^2}{4 R^2}(2 \sqrt{2}+1)=\frac{M V^2}{R} \text { or } V=\sqrt{\frac{G M}{R}\left(\frac{2 \sqrt{2}+1}{4}\right)}
\)

Hint

Acceleration due to gravity at a height \(=\frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})^2}=\frac{6.67 \times 10^{-11} \times 7.4 \times 10^{22}}{(1740+1000)^2 \times 10^6}\)
\(
\begin{aligned}
&=\frac{49.358 \times 10^{11}}{2740 \times 2740 \times 10^6} \\
&=\frac{49.358 \times 10^{11}}{0.75 \times 10^{13}}=65.3 \times 10^{-2}=0.65 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)

Hint

The linear momentum of 2 bodies is o initially. Since gravitational force is internal, final momentum is also zero
So \((10 \mathrm{~kg}) \mathrm{v}_1=(20 \mathrm{~kg}) \mathrm{v}_2\)
Or
\(
\mathrm{v}_1=\mathrm{v}_2
\)
Since P.E conserved
initial P.E \(=\frac{-6.67 \times 10^{-11} \times 10 \times 20}{1}=-13.34 \times 10^{-9} \mathrm{~J}\)
When separation is \(0.5 \mathrm{~m}\),
\(
\Rightarrow-13.34 \times 10^{-9}+0=\frac{-13.34 \times 10^{-9}}{(1 / 2)}+(1 / 2) \times 10 \mathrm{v}_1^2+(1 / 2) \times 20 \mathrm{v}_2^2
\)
\(
\Rightarrow-13.34 \times 10^{-9}=26.68 \times 10^{-9}+5 \mathrm{v}_1^2+10 \mathrm{v}_2^2
\)
\(
\Rightarrow-13.34 \times 10^{-9}=-26.68 \times 10^{-9}+30 \mathrm{v}_2^2
\)
\(
\Rightarrow \mathrm{v}_2^2=\frac{13.34 \times 10^{-9}}{30}=4.44 \times 10^{-10}
\)
\(
\Rightarrow \mathrm{v}_2=2.1 \times 10^{-5} \mathrm{~m} / \mathrm{s}
\)
\(
\text { So, } \mathrm{v}_1=4.2 \times 10^{-5} \mathrm{~m} / \mathrm{s}
\)

Hint

Consider a small mass element of length \(d\) subtending \(d \theta\) angle at the centre.

In the semicircle, we can consider a small element \(d \theta\).
Then length of the element, \(d=R d \theta\)
Mass of the element, \(\mathrm{dm}=\left(\frac{M}{L}\right) R d \theta\)
Force on the mass element is given by
\(
d F=\frac{G m}{R^2} d m=\frac{G M R m}{L R^2} d \theta
\)
The symmetric components along \(A B\) cancel each other.
Now, net gravitational force on the particle at \(O\) is given by
\(
\begin{aligned}
&F=\int 2 d F \sin \theta \\
&=\int \frac{2 G M m}{L R} \sin \theta d \theta \\
&\therefore F=\int_0^\pi / 2 \frac{-2 G M m}{L R} \sin \theta d \theta \\
&=\frac{2 G M m}{L R}[-\cos \theta]_0^\pi / 2 \\
&=-2 \frac{G M m}{L R}(-1) \\
&=\frac{2 G M m}{L R}=\frac{2 G M m}{L L / \pi} \\
&=\frac{2 \pi G M m}{L^2}
\end{aligned}
\)

Hint

Consider a small mass element of length \(\mathrm{d} x\) at a distance \(\mathrm{x}\) from the centre of the rod. Mass of the mass element, \(d m=(M / L) \times d x\)
\(
\text { Gravitational field due to this element at point } \mathrm{P} \text { is given by } d E_1=\frac{G(d m) \times 1}{\left(d^2+x^2\right)}= d E_2
\)
The components of the gravitational field due to the symmetrical mass element along the length of the rod cancel each other.
Now, resultant gravitational field \(=2 \mathrm{dE_1} \sin \theta\)
\(
\begin{aligned}
&=2 \times \frac{G(d m)}{\left(d^2+x^2\right)} \times \frac{d}{\sqrt{\left(d^2+x^2\right)}} \\
&=\frac{2 \times G M \times d d x}{L\left(d^2+x^2\right)\left\{\left(\sqrt{d^2+x^2}\right)\right\}}
\end{aligned}
\)
Total gravitational field due to the rod at point \(\mathrm{P}\) is given by \(E=\int_0^{L / 2} \frac{2 G m d d x}{L\left(d^2+x^2\right)^{3 / 2}}\)
On integrating the above equation, we get:
\(
E=\frac{2 G m}{d \sqrt{L^2+4 d^2}}
\)

Hint

Field due to a spherical shell at an inside point is zero. Hence the field at the point \(P\) will be only due to \(M_1\).
So, the gravitational field at point \(\mathrm{P}\),
\(
E_g=\frac{4 G M_1}{\left(R_1+R_2\right)^2}
\)
Thus, the force on mass \(m\) is given by:
\(
F=m E_g
\)
Substituting the value of \(E_g\),
\(
\therefore F=\frac{4 G M_1 m}{\left(R_1+R_2\right)^2}
\)

Hint

Mass of the Earth,
\(
M=\left(\frac{4}{3}\right) \pi R^3 \rho \text {…(i) }
\)
Consider an imaginary sphere of radius \(\mathrm{x}\) with centre \(\mathrm{O}\) as shown in the figure above
Mass of the imaginary sphere,\(M^{\prime}=\left(\frac{4}{3}\right) \pi x^3 \rho \ldots(\) ii \()\)
From (i) and (ii), we have:
\(
\frac{M^{\prime}}{M}=\frac{x^3}{R^3}
\)
\(\therefore\) Gravitational force on the particle of mass \(\mathrm{m}\) is given by \(\mathrm{F}=\frac{G M^{\prime} m}{x^2}\)
\(
\Rightarrow F=\frac{G M x^3 m}{R^3 x^2}=\frac{G M m}{R^3} x
\)

Hint

Let d be the distance from centre of earth to mass ‘m’ then \(d=\sqrt{x^2+\left(\frac{R^2}{4}\right)}=(1 / 2) \sqrt{4 x^2+R^2}\)
\(M\) be the mass of the earth, \(M^{\prime}\) the mass of the sphere of radius \(d / 2\).
Then \(M=(4 / 3) \pi R^3 \rho\)
\(
\mathrm{M}^{\prime}=(4 / 3) \pi \mathrm{d}^3 \rho
\)
or \(\frac{M^{\prime}}{M}=\frac{d^3}{R^3}\)
\(\therefore\) Gravitational force exerted by the earth on the particle of mass \(m\),
\(
F=\frac{G M^{\prime} m}{d^2}=\frac{G d^3 M m}{R^3 d^2}=\frac{G M m d}{R^3}
\)
So, Normal force exerted by the wall \(=\mathrm{F} \cos \theta\).
\(=\frac{G M m d}{R^3} \times \frac{R}{2 d}=\frac{G M m}{2 R^2} \quad\)

Hint

(a) Consider that the particle is placed at a distance \(x\) from \(O\).
Here, \(r<x<2 r\)
Let us consider a thin solid sphere of radius \((x-r)\).
Let’s consider a thin shell of mass
\(
d m=\frac{m}{\left(\frac{4}{3}\right) \pi r^3} \times \frac{4}{3} \pi(x-r)^3=\frac{m(x-r)^3}{r^3}
\)
Then the gravitational force on the particle due to the solid sphere is given by
\(
F=\frac{G m^{\prime} d m}{(x-r)^2}
\)
\(
=\frac{G \frac{m(x-r)^3}{r^3} m^{\prime}}{(x-r)^2}=\frac{G m m^{\prime}(x-r)}{r^3}
\)
\(
\text { Force on the particle due to the shell will be zero because gravitational field intensity inside a shell is zero. }
\)

(b) If \(2 \mathrm{r}<\mathrm{x}<2 \mathrm{R}\),
Force on the body due to the shell will again be zero as particle is still inside the shell. then \(\mathrm{F}\) is only due to the solid sphere.
\(
\therefore F=\frac{G m m^{\prime}}{(x-r)^2}
\)

(c) If \(x>2 R\), then the gravitational force is due to both the sphere and the shell.
Now, we have:
Gravitational force due to shell,
\(
F=\frac{G M m^{\prime}}{(x-R)^2}
\)
Gravitational force due to the sphere \(=\frac{G m m^{\prime}}{(x-r)^2}\)
As both the forces are acting along the same line joining the particle with the centre of the sphere and shell so both the forces can be added directly without worrying about their vector nature.
\(\therefore\) Resultant force \(=\frac{G m m^{\prime}}{(x-r)^2}+\frac{G M m^{\prime}}{(x-R)^2}\)

Hint

At \(\mathrm{P}_1\), Gravitational field due to sphere \(M=\frac{\mathrm{GM}}{(3 \mathrm{a}+\mathrm{a})^2}=\frac{\mathrm{GM}}{16 \mathrm{a}^2}\) At \(\mathrm{P}_2\), Gravitational field is due to sphere \& shell,
\(
=\frac{\mathrm{GM}}{(\mathrm{a}+4 \mathrm{a}+\mathrm{a})^2}+\frac{\mathrm{GM}}{(4 \mathrm{a}+\mathrm{a})^2}=\frac{\mathrm{GM}}{\mathrm{a}^2}\left(\frac{1}{36}+\frac{1}{25}\right)=\left(\frac{61}{900}\right) \frac{\mathrm{GM}}{\mathrm{a}^2}
\)

Hint

We know in the thin spherical shell of uniform density has gravitational field at its internal point is zero. At A and B point, field is equal and opposite and cancel each other so the Net field is zero.
Hence, \(E_A=E_B\)

Hint

Let the mass of \(0.10 \mathrm{~kg}\) be at a distance \(x\) from the \(2 \mathrm{~kg}\) mass and at a distance of \((2-\mathrm{x})\) from the \(4 \mathrm{~kg}\) mass. Force between \(0.1 \mathrm{~kg}\) mass and \(2 \mathrm{~kg}\) mass = force between \(0.1 \mathrm{~kg}\) mass and \(4 \mathrm{~kg}\) mass \(\therefore \frac{2 \times 0.1}{x^2}=-\frac{4 \times 0.1}{(2-x)^2}\)
\(
\begin{aligned}
&\Rightarrow \frac{0.2}{x^2}=\frac{0.4}{(2-x)^2} \\
&\Rightarrow \frac{1}{x^2}=\frac{2}{(2-x)^2} \\
&\Rightarrow(2-x)^2=2 x^2 \\
&\Rightarrow 2-x=\sqrt{2} x \\
&\Rightarrow x(\sqrt{2}+1)=2 \\
&\Rightarrow x=\frac{2}{2.414}
\end{aligned}
\)
\(=0.83 \mathrm{~m}\) from the \(2 \mathrm{~kg}\) mass
Now, gravitation potential energy of the system is given by
\(
\begin{aligned}
&U=-G\left[\frac{0.1 \times 2}{0.83}+\frac{0.1 \times 4}{1.17}+\frac{2 \times 4}{2}\right] \\
&\Rightarrow U=-6.67 \times 10^{11}\left[\frac{0.1 \times 2}{0.83}+\frac{0.1 \times 4}{1.17}+\frac{2 \times 4}{2}\right] \\
&\Rightarrow U=-3.06 \times 10^{-10} \mathrm{~J}
\end{aligned}
\)

Hint

Initially, the side of \(\Delta\) is a. To increase it to \(2 \mathrm{a}\),
work done \(=\frac{G m^2}{2 a}+\frac{G m^2}{a}=\frac{3 G m^2}{2 a}\)

Hint

Initial distance of the body from the centre of the sphere \(=10 \mathrm{~cm}=0.1 \mathrm{~m}=d_1=10^{-1} \mathrm{~m}\)
and final distance of the body \(=\infty=d_2\)
\(\therefore\) The work done in taking a body far away from the sphere against the gravitational force between them
\(=\) Change in gravitational potential energy
i.e. \(U_2-U_1\)
\(=-\frac{G M m}{d_2}-\left(-\frac{G M m}{d_1}\right)\)
\(=-\frac{G M m}{\infty}+\frac{G M m}{0.1}=0+\frac{G M m}{0.1}\)
\(\therefore U_2-U_1=\frac{G M m}{0.1}\)
\(=\frac{6.67 \times 10^{-11} \times 100 \times 10 \times 10^{-3}}{10^{-1}}[latex] [latex]=6.67 \times 10^{-10} \mathrm{~J}\)

Hint

\(\vec{E}=(5 \mathrm{~N} / \mathrm{kg}) \hat{\mathrm{i}}+(12 \mathrm{~N} / \mathrm{kg}) \hat{\mathrm{j}}\)
a) \(\vec{F}=\vec{E} m\)
\(=2 \mathrm{~kg}[(5 \mathrm{~N} / \mathrm{kg}) \hat{i}+(12 \mathrm{~N} / \mathrm{kg}) \hat{j}]=(10 \mathrm{~N}) \hat{i}+(12 \mathrm{~N}) \hat{\mathrm{j}}\)
\(|\vec{F}|=\sqrt{100+576}=26 \mathrm{~N}\)

(b) \(V=-\vec{E} \cdot \vec{r}\)
Potential at \((12 \mathrm{~m}, 0)=-60 J / K g\)
Potential at \((0,5 \mathrm{~m})=-60 \mathrm{~J} / \mathrm{kg}\)

(c)
change in potential= final potential -initial potential
initial potential \(=\) potential at the origin \(=0\) final potential \(=\) potential at \((12,5)\)
\(
\begin{aligned}
&V=-\vec{E} \cdot \vec{r} \\
&=-(10 \hat{i}+24 \widehat{j}) \cdot(12 \hat{i}+5 \widehat{j}) \\
&=-(120+120) J \\
&=-240 \mathrm{~J}
\end{aligned}
\)

(d)
\(
\begin{aligned}
&\Delta V=-\vec{E} \cdot \Delta \vec{r} \\
&\Delta \vec{r}=(12 \widehat{i}+0 \widehat{j})-(0 \widehat{i}+5 \widehat{j}) \\
&=12 \widehat{i}-5 \widehat{j} \\
&\Delta V=-(10 \widehat{i}+24 \widehat{j}) \cdot(12 \widehat{i}-5 \widehat{j}) \\
&=0 J
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) } V=\left(\frac{20 N}{k g}\right)(x+y) \\
&{\left[\frac{G M}{R}\right]=\left[\frac{M L T^{-2}}{M}\right][L]} \\
&\Rightarrow\left[\frac{M^{-1} L^3 T^{-2} M^1}{L}\right]=\left[\frac{M L^2 T^{-2}}{M}\right] \\
&\Rightarrow\left[M^0 L^2 T^{-2}\right]=\left[M^0 L^2 T^{-2}\right] \\
&\therefore L H S=R H S
\end{aligned}
\)
(b) The gravitational field at the point \((x, y)\) is given by \(\vec{E}_{(x, y)}=-20\left(\frac{N}{k g}\right) \hat{i}-\left(\frac{20 N}{k g}\right) \hat{j}\)
\(
\begin{aligned}
&\text { (c) } \vec{F}=\vec{E} m \\
&=0.5 k g\left[-\left(\frac{20 N}{k g}\right) \hat{i}-\left(\frac{20 N}{k g}\right) \hat{j}\right] \\
&=-(10 N) \hat{i}-(10 N) \hat{j} \\
&\therefore|\vec{F}|=\sqrt{(100)+(100)} \\
&=10 \sqrt{2} N
\end{aligned}
\)

Hint

The gravitational field in a region is given by \(\vec{E}=2 \hat{i}+3 \hat{j}\)
Slope of the electric field, \(m_1=\tan \theta_1=\frac{3}{2}\)
The given line is \(3 y+2 x=5\).
Slope of the line, \(m_2=\tan \theta_2=-\frac{2}{3}\)
We can see that \(m_1 m_2=-1\)
Since the directions of the field and the displacement are perpendicular to earth other, no work is done by the gravitational field when a particle is moved on the given line.

Hint

Assume that at height h, the weight of the body becomes half.
Weight of the body at the surface \(=m g\)
Weight of the body at height \(h\) above the Earth’s surface \(=\mathrm{mg}^{\prime}\), where \(\mathrm{g}^{\prime}\) is the acceleration due to gravity at height \(h\)
Now,\(g^{\prime}=\frac{1}{2} g\)
\(
\begin{aligned}
&\therefore\left(\frac{1}{2}\right) \frac{G M}{R^2}=\frac{G M}{(R+h)^2}\left[\because g=\frac{G M}{R^2}\right] \\
&\Rightarrow 2 R^2=(R+h)^2 \\
&\Rightarrow \sqrt{2} R=R+h \\
&\Rightarrow h=(\sqrt{2}-1) R
\end{aligned}
\)

Hint

Let \(\mathrm{g}^{\prime}\) be the acceleration due to gravity on mount Everest.
\(
\begin{aligned}
&g^{\prime}=g\left(1-\frac{2 \mathrm{~h}}{\mathrm{R}}\right) \\
&=9.8\left(1-\frac{17696}{6400000}\right)=9.8(1-0.00276)=9.77 \mathrm{~m} / \mathrm{s}^2
\end{aligned}
\)

Hint

Let \(\mathrm{g}^{\prime}\) be the acceleration due to gravity in mine.
Then \(g^{\prime}=g\left(1-\frac{d}{R}\right)\)
\(
=9.8\left(1-\frac{640}{6400 \times 10^3}\right)=9.8 \times 0.9999=9.799 \mathrm{~m} / \mathrm{s}^2
\)

Hint

Let \(g^{\prime}\) be the acceleration due to gravity at equator & that of pole \(=g\)
\(
\begin{aligned}
&\mathrm{g}^{\prime}=\mathrm{g}-\omega^2 \mathrm{R} \\
&=9.81-\left(7.3 \times 10^{-5}\right)^2 \times 6400 \times 10^3 \\
&=9.81-0.034 \\
&=9.776 \mathrm{~m} / \mathrm{s}^2 \\
&\mathrm{mg}^{\prime}=1 \mathrm{~kg} \times 9.776 \mathrm{~m} / \mathrm{s}^2 \\
&=9.776 \mathrm{~N} \text { or } 0.997 \mathrm{~kg}
\end{aligned}
\)
The body will weigh \(0.997 \mathrm{~kg}\) at equator.

Hint

At the equator, \(g^{\prime}=g-\omega^2 R \quad \quad\)…(i)
Let \(h\) be the height above the South Pole where the body stretch the spring by the same length. Let \(\mathrm{h}\) be the height above the South Pole where the body stretch The acceleration due to gravity at this point is \(g^{\prime}=g\left(1-\frac{2 h}{R}\right)\)
Weight of the body at the equator = weight of the body at height \(h\) above the South Pole
\(
\begin{aligned}
&\therefore g-\omega^2 r=g\left(1-\left(\frac{2 h}{R}\right)\right) \\
&\Rightarrow 1-\frac{\omega^2 R}{2 g}=1-\frac{2 h}{R} \\
&\Rightarrow h=\frac{\omega^2 R^2}{2 g} \\
&=\frac{\left(7.3 \times 10^{-5}\right)^2 \times\left(6400 \times 10^{-3}\right)^2}{2 \times 9.81} \\
&=\frac{(7.3)^2 \times(64)^2}{19.62} \\
&=11125 \mathrm{~m}=10 \mathrm{~km}(\text { approx. })
\end{aligned}
\)

Hint

The apparent acceleration due to gravity at the equator becomes zero.
\(
\begin{aligned}
&\text { i.e., } g^{\prime}=g-\omega^2 R=0 \\
&\Rightarrow g=\omega^2 R
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow \omega=\sqrt{\frac{g}{R}}=\sqrt{\frac{9.8}{6400 \times 10^3}} \\
&\Rightarrow \omega=\sqrt{\frac{9.8 \times 10^{-5}}{6.4}} \\
&\Rightarrow \omega=1.237 \times 10^{-3} \mathrm{rad} / \mathrm{s} \\
&\therefore T=\frac{2 \pi}{\omega}=\frac{2 \times 3.14}{1.2 \times 10^{-3}} \\
&=\frac{6.28}{1.2 \times 10^{-3}}
\end{aligned}
\)
\(
=1.41 \mathrm{~h}
\)

Hint

(a) Speed of the ship due to rotation of the Earth is \(V=\omega R\), where \(R\) is the radius of the Earth and \(\omega\) is its angular speed.
(b) The tension in the string is given by
\(
\begin{aligned}
&T_0=m g+m \omega^2 R \\
&\therefore T_0-m g=m \omega^2 R
\end{aligned}
\)

(c)

\(
\begin{aligned}
&v=\omega_1 R \\
&V=\omega R
\end{aligned}
\)
Then the tension in the string is given by
\(
\begin{array}{lll}
T=m g+m\left(\omega_1+\omega\right)^2 R \dots(i) \\
T_0=m g+m \omega^2 R \dots(ii)
\end{array}
\)
\(
\begin{aligned}
T-T_0 &=m R\left[\left(\omega_1+\omega\right)^2-\omega^2\right] \\
&=m R\left(\omega_1^2+\omega^2+2 \omega \omega_1-\omega^2\right) \\
T-T_0 &=m R\left(\omega_1^2+2 \omega_1 \omega\right)
\end{aligned}
\)
\(
\omega_1=\frac{v}{R}
\)
As \(R\) is big \(\omega_1^2\) is negligible and we can ignore it.
\(
\begin{aligned}
&T-T_0=m R(2 \omega_1 \omega) \\
&T-T_0=2 \omega m v \\
&T=T_0+2 \omega m v \\
\end{aligned}
\)

Hint

According to Kepler’s laws of planetary motion, the time period of revolution of a planet about the Sun is directly proportional to the cube of the distance between their centres.
i.e., \(T^2 \propto R^3\)
\(
\begin{aligned}
&\Rightarrow \frac{T_m^2}{T_e^2}=\frac{R_m^3}{R_e^3} \\
&\Rightarrow\left(\frac{R_m}{R_e}\right)^3=\left(\frac{1.88}{1}\right)^2
\end{aligned}
\)
\(
\therefore \frac{R_m}{R_e}=(1.88)^{2 / 3}=1.52
\)

Hint

\(
\begin{aligned}
&T=2 \pi \sqrt{\frac{r^3}{G M}} \\
&27.3=2 \times 3.14 \sqrt{\frac{\left(3.84 \times 10^5\right)^3}{6.67 \times 10^{-11} \times M}} \\
&\text { or } 2.73 \times 2.73=\frac{2 \times 3.14 \times\left(3.84 \times 10^5\right)^3}{6.67 \times 10^{-11} \times M} \\
&\text { or } M=\frac{2 \times(3.14)^2 \times(3.84)^3 \times 10^{15}}{3.335 \times 10^{-11}(27.3)^2}=6.02 \times 10^{24} \mathrm{~kg}
\end{aligned}
\)
\(\therefore\) mass of earth is found to be \(6.02 \times 10^{24} \mathrm{~kg}\).

Hint

Time period of revolution of the satellite around the Mars is give by \(T=2 \pi \sqrt{\frac{r^3}{G M}}\) where \(\mathrm{M}\) is the mass of the Mars and \(r\) is the distance of the satellite from the centre of the planet.
\(
\begin{aligned}
&\text { Now }, 27540=2 \times 3.14 \sqrt{\frac{\left(9.4 \times 10^3 \times 10^3\right)^3}{6.67 \times 10^{-11} \times M}} \\
&\Rightarrow(27540)^2=(6.28)^2 \times \frac{\left(9.4 \times 10^5\right)^3}{6.67 \times 10^{-11} \times M} \\
&\Rightarrow M=\frac{(6.28)^2 \times(9.4)^3 \times 10^{18}}{6.67 \times 10^{-11} \times(27540)^2} \\
&\Rightarrow M=6.5 \times 10^{23} \mathrm{~kg}
\end{aligned}
\)

Hint

(a) Speed of the satellite in its orbit \(v=\sqrt{\frac{G M}{r+h}}=\sqrt{\frac{g r^2}{r+h}}\)
\(
\begin{aligned}
&\Rightarrow v=\sqrt{\frac{9.8 \times\left(6400 \times 10^3\right)^2}{10^6 \times(6.4+2)}} \\
&\Rightarrow v=\sqrt{\frac{9.8 \times 6.4 \times 6.4 \times 10^6}{8.4}} \\
&\Rightarrow v=6.9 \times 10^3 \mathrm{~m} / \mathrm{s}=6.9 \mathrm{~km} / \mathrm{s}
\end{aligned}
\)
(b) Kinetic energy of the satellite
\(
\begin{aligned}
&K . E .=\frac{1}{2} m v^2 \\
&=\frac{1}{2} \times 1000 \times\left(6.9 \times 10^3\right)^2 \\
&=\frac{1}{2} \times 1000 \times\left(47.6 \times 10^6\right) \\
&=2.38 \times 10^{10} \mathrm{~J}
\end{aligned}
\)
(c) Potential energy of the satellite
\(
\begin{aligned}
&P . E .=-\frac{G M m}{(R+h)} \\
&=-\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 10^3}{(6400+2000) \times 10^3} \\
&=\frac{40 \times 10^{13}}{8400}=-4.76 \times 10^{10} \mathrm{~J}
\end{aligned}
\)
(d) Time period of the satellite \(T=\frac{2 \pi(r+h)}{v}\)
\(
\begin{aligned}
&=\frac{2 \times 3.14 \times 8400 \times 10^3}{6.9 \times 10^3} \\
&=\frac{6.28 \times 84 \times 10^2}{6.9} \\
&=76.6 \times 10.2 \mathrm{~s} \\
&=2.1 \mathrm{~h}
\end{aligned}
\)

Hint

(a) The angular speed of the Earth and the satellite will be the same.
i.e., \(\frac{2 \pi}{T_e}=\frac{2 \pi}{T_s}\)
\(
\begin{aligned}
&\Rightarrow \frac{1}{24 \times 3600}=\frac{1}{2 \pi \sqrt{(R+h)^3 / g h^2}} \\
&\Rightarrow 12 \times 3600=3.14 \sqrt{\frac{(R+h)^3}{g R^2}} \\
&\Rightarrow \frac{(R+h)^3}{g R^2}=\frac{(12 \times 3600)^2}{(3.14)^2} \\
&\Rightarrow \frac{\left(6400+h^3\right) \times 10^9}{9.8 \times(6400)^2 \times 10^6}=\frac{(12 \times 3600)^2}{(3.14)^2} \\
&\Rightarrow \frac{(6400+h) \times 10^9}{6272 \times 10^9}=432 \times 10^4 \\
&\Rightarrow(6400+h)^3=6272 \times 432 \times 10^4 \\
&\Rightarrow 6400+h=\left(6272 \times 432 \times 10^4\right)^{1 / 3}-6400 \\
&\Rightarrow h=42300 \mathrm{~km}
\end{aligned}
\)
(b) Time taken from the North Pole to the equatorial plane is given by \(\frac{1}{4} T\)
\(
\begin{aligned}
&=\frac{1}{4} \times 6.28 \sqrt{\frac{(42300+6400)^3}{10 \times(6400)^2 \times 10^6}} \\
&=3.14 \sqrt{\frac{(479)^3 \times 10^6}{(64)^2 \times 10^{11}}} \\
&=3.14 \sqrt{\frac{497 \times 497 \times 497}{64 \times 64 \times 10^5}} \\
&=6 h
\end{aligned}
\)

Hint

For a geostationary satellite, we have:
\(
\begin{aligned}
&\mathrm{R}=6.4 \times 10^3 \mathrm{~km} \\
&\mathrm{~h}=3.6 \times 10^3 \mathrm{~km} \\
&\text { Given: } \mathrm{mg}=10 \mathrm{~N}
\end{aligned}
\)
The true weight of the object in the geostationary satellite is given by
\(
\begin{aligned}
&\mathrm{mg}^{\prime}=\mathrm{mg}-\frac{R^2}{(R+h)^2} \\
&=10-\frac{\left(6400 \times 10^3\right)^2}{\left(6400 \times 10^3+3600 \times 10^3\right)} \\
&=10-\left[\frac{\left(64 \times 10^5\right)^2}{\left(6.4 \times 10^6+36 \times 10^5\right)}\right] \\
&=10-\left[\frac{4096 \times 10^{10}}{(42.4)^2 \times 10^{12}}\right] \\
&=\frac{4096}{17980}=0.23 N
\end{aligned}
\)

Hint

The time period of revolution of the satellite around a planet in terms of the radius of the planet and radius of the orbit of the satellite is given by \(T=2 \pi \sqrt{\frac{R_2^3}{g R_1^2}}\), where \(\mathrm{g}\) is the acceleration due to gravity at the surface of the planet.
\(
\begin{aligned}
&\text { Now }, T^2=4 \pi^2 \frac{R_2^3}{g R_1^2} \\
&\Rightarrow g=\frac{4 \pi^2}{T^2} \frac{R_2^3}{R_1^2}
\end{aligned}
\)
\(\therefore\) Acceleration due to gravity of the planet \(=\frac{4 \pi^2}{T^2} \frac{R_2^3}{R_1^2}\)

Hint

The colattitude is given by \(\phi\).
\(
\angle \mathrm{OAB}=90^{\circ}-\angle \mathrm{ABO}
\)
Again \(\angle \mathrm{OBC}=\phi=\angle \mathrm{OAB}\)
\(
\begin{aligned}
&\therefore \sin \phi=\frac{6400}{42000}=\frac{8}{53} \\
&\therefore \phi=\sin ^{-1}\left(\frac{8}{53}\right)=\sin ^{-1} 0.15
\end{aligned}
\)

Hint

The particle attains maximum height \(=6400 \mathrm{~km}\). On earth’s surface, its P.E. & K.E.
\(
E_e=(1 / 2) m v^2+\left(\frac{-G M m}{R}\right) \dots(1)
\)
In space, its P.E. & K.E.
\(
\begin{aligned}
&E_s=\left(-\frac{G M m}{R+h}\right)+0 \\
&E_s=\left(-\frac{G M m}{2 R}\right) \dots(2)
\end{aligned}
\)\(\quad(\because h=R)\)
Equating (1) & (2)
\(
-\frac{G M m}{R}+\frac{1}{2} m v^2=-\frac{G M m}{2 R}
\)
\(\operatorname{Or}(1 / 2) m v^2=G M m\left(-\frac{1}{2 R}+\frac{1}{R}\right)\)
\(
\begin{aligned}
&\text { Or } v^2=\frac{\mathrm{GM}}{\mathrm{R}} \\
&=\frac{6.67 \times 10^{-11} \times 6 \times 10^{24}}{6400 \times 10^3} \\
&=\frac{40.02 \times 10^{13}}{6.4 \times 10^6} \\
&=6.2 \times 10^7=0.62 \times 10^8 \\
&\text { Or } v=\sqrt{0.62 \times 10^8}=0.79 \times 10^4 \mathrm{~m} / \mathrm{s}=7.9 \mathrm{~km} / \mathrm{s}
\end{aligned}
\)

Hint

Initial velocity of the particle, \(\mathrm{v}=15 \mathrm{~km} / \mathrm{s}\)
Let its speed be \(v^{\prime}\) in interstellar space.
Applying the law of conservation of energy, we have:
\(
\begin{aligned}
&\left(\frac{1}{2}\right) m\left[v-v^{\prime 2}\right]=\int_R^{\infty} \frac{G M m}{x^2} d x \\
&\therefore\left(\frac{1}{2}\right) m\left[15 \times 10^3-v^{\prime 2}\right]=\int_R^{\infty} \frac{G M m}{x^2} d x \\
&\Rightarrow\left(\frac{1}{2}\right) m\left[\left(15 \times 10^3\right)^2-v^{\prime 2}\right]=G M m\left[\frac{-1}{x}\right] \\
&\Rightarrow\left(\frac{1}{2}\right) m\left[\left(225 \times 10^5\right)-v^{\prime 2}\right]=\frac{G M m}{R} \\
&\Rightarrow 225 \times 10^5-v^{\prime 2}=\frac{2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{6400 \times 10^3} \\
&\Rightarrow v^{\prime 2}=225 \times 10^6-\frac{40.02}{32} \times 10^8 \\
&=2.25 \times 10^8-1.2 \times 10^8 \\
&=10^8(1.05) \\
&\text { Or } v^{\prime}=1.01 \times 10^4 \mathrm{~m} / \mathrm{s}=10 \mathrm{~km} / \mathrm{s}
\end{aligned}
\)

Hint

Mass of the sphere \(=6 \times 10^{24} \mathrm{~kg}\)
Escape velocity \(=3 \times 10^8 \mathrm{~m} / \mathrm{s}\)
Escape velocity is given by
\(
\begin{aligned}
&v_e=\frac{2 G M}{R} \\
&\Rightarrow R=\frac{2 G M}{v_e^2} \\
&=\frac{2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24}}{\left(3 \times 10^8\right)^2} \\
&=\frac{2 \times 40.02 \times 10^{13}}{9 \times 10^{16}} \\
&=\frac{80.02}{9} \times 10^{-3} \mathrm{~m} \\
&=8.89 \times 10^{-3} \mathrm{~m} \\
& \approx 9 \mathrm{~mm}
\end{aligned}
\)