JEE PYQs Integer Type

Hint

(d) Step 1: Establish the scaling relationship
The time \(\boldsymbol{T}\) for three identical spheres initially at rest at the vertices of an equilateral triangle to collapse and collide under their mutual gravitational force scales with the initial side length \(a\) and mass \(m\) as \(T \propto \frac{a^{3 / 2}}{\sqrt{m}}\). This can be expressed as:
\(
\frac{T_2}{T_1}=\frac{a_2^{3 / 2} / \sqrt{m_2}}{a_1^{3 / 2} / \sqrt{m_1}}
\)
Step 2: Substitute the given values
We are given the initial time \(T_1=4 \mathrm{~s}\), initial side length \(a_1=a\), and initial mass \(m_1=m\) . The new parameters are \(a_2=2 a\) and \(m_2=2 m\). Substituting these into the ratio:
\(
\frac{T_2}{4 \mathrm{~s}}=\frac{(2 a)^{3 / 2} / \sqrt{2 m}}{a^{3 / 2} / \sqrt{m}}
\)
Step 3: Simplify and calculate the new time
Simplify the expression:
\(
\frac{T_2}{4 \mathrm{~s}}=\frac{2^{3 / 2} a^{3 / 2} /(\sqrt{2} \sqrt{m})}{a^{3 / 2} / \sqrt{m}}=\frac{2^{3 / 2}}{\sqrt{2}}=\frac{2 \sqrt{2}}{\sqrt{2}}=2
\)
So, \(\frac{T_2}{4 \mathrm{~s}}=2\).
Solving for \(T_2\) :
\(
T_2=2 \times 4 \mathrm{~s}=8 \mathrm{~s}
\)

Explanation: Dimensional Analysis (The Intuitive Way):
The time \(T\) it takes for the spheres to collide depends on three physical quantities:
1. The initial distance (\(a\)).
2. The mass of the spheres ( \(m\) ).
3. The Gravitational Constant \((G)\), which dictates the strength of the force.
We can say \(T=k \cdot a^x \cdot m^y \cdot G^z\). By matching the dimensions of time ( \(\left[M^0 L^0 T^1\right]\) ) on both sides:
\([G]=M^{-1} L^3 T^{-2}\)
\([m]=M\)
\([a]=L\)
If you solve for the exponents, you find that to get “Time” on the right side, the relationship must be:
\(
T \propto \sqrt{\frac{a^3}{G m}}
\)
Equation of Motion (The Rigorous Way):
If you look at one sphere, the net gravitational force \(F\) pulling it toward the center of the triangle is the vector sum of the forces from the other two spheres.
The magnitude of force from one sphere is \(F_g=\frac{G m m}{a^2}\). The component of this force directed toward the center of the equilateral triangle results in an acceleration:
\(
a_{\text {accel }}=\frac{F_{\text {net }}}{m} \propto \frac{G m}{a^2}
\)
From basic kinematics, we know that distance is roughly proportional to acceleration multiplied by time squared \(\left(s \approx \frac{1}{2} a t^2\right)\). Substituting our variables:
\(
a \propto\left(\frac{G m}{a^2}\right) T^2
\)
Rearranging for \(T^2\) :
\(
\begin{aligned}
T^2 & \propto \frac{a^3}{G m} \\
T & \propto \sqrt{\frac{a^3}{G m}}
\end{aligned}
\)
In your specific problem, \(G\) is a constant, so we simplify the proportionality to:
\(
T \propto \sqrt{\frac{a^3}{m}}
\)

Hint

(c) To find the kinetic energy ( \(K E\) ) of a satellite in a stable circular orbit, we use the relationship between gravitational force and centripetal force.
The Formula:
For a satellite of mass \(m\) orbiting a planet of mass \(M\) at a distance \(r\) from the center, the orbital velocity \(v\) is given by:
\(
v=\sqrt{\frac{G M}{r}}
\)
The formula for Kinetic Energy is \(K E=\frac{1}{2} m v^2\). Substituting the orbital velocity, we get:
\(
K E=\frac{G M m}{2 r}
\)
Identify the Variables:
Mass of Earth (M): \(6 \times 10^{24} \mathrm{~kg}\)
Mass of Satellite (\(m\)): 1000 kg (or \(10^3 \mathrm{~kg}\) )
Gravitational Constant (\(G\)): \(6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{~kg}^{-2}\)
Radius of Earth \((R)\) : \(6.4 \times 10^6 \mathrm{~m}\)
Height (h): \(270 \mathrm{~km}=0.27 \times 10^6 \mathrm{~m}\)
Orbital Radius \((r): R+h=6.4 \times 10^6+0.27 \times 10^6=6.67 \times 10^6 \mathrm{~m}\)
Calculation:
Plug the values into the \(K E\) formula:
\(
K E=\frac{\left(6.67 \times 10^{-11}\right) \times\left(6 \times 10^{24}\right) \times\left(10^3\right)}{2 \times\left(6.67 \times 10^6\right)}
\)
Notice that the value 6.67 appears in both the numerator and the denominator, so they cancel out nicely:
\(
\begin{gathered}
K E=\frac{6 \times 10^{-11+24+3}}{2 \times 10^6} \\
K E=\frac{6 \times 10^{16}}{2 \times 10^6} \\
K E=3 \times 10^{10} \mathrm{~J}
\end{gathered}
\)
The kinetic energy of the satellite is \(3 \times 10^{10} \mathrm{~J}\).

Hint

(b)

The Formula for Angular Momentum:
For a planet of mass \(m\) moving in a circular orbit of radius \(r\) with orbital velocity \(v\), the angular momentum \(L\) is:
\(
L=m v r
\)
Orbital velocity:
The angular momentum \(L\) of a planet with mass \(m\) in a circular orbit of radius \(r\) is given by \(L=m v r\), where \(v\) is the orbital speed. The orbital speed is determined by balancing the gravitational force and the centripetal force:
\(
\frac{m v^2}{r}=G \frac{{M_s} m}{r^2}
\)
Solving for \(v\), we get:
\(
v=\sqrt{\frac{G M_s}{r}}
\)
Substituting this into the angular momentum formula:
\(
\begin{gathered}
L=m\left(\sqrt{\frac{G M_s}{r}}\right) r \\
L=m \sqrt{G M_s r}
\end{gathered}
\)
Since \(G\) and \(M_s\) (the mass of the star) are constant for both planets, we find the proportionality:
\(
L \propto m \sqrt{r}
\)
Setting up the Ratio:
We are given the following relationships between Planet B and Planet A:
Mass: \(m_B=4 \sqrt{2} m_A\)
Radius: \(R_B=2 R_A\)
Now, we calculate the ratio \(\frac{L_B}{L_A}\) :
\(
\frac{L_B}{L_A}=\left(\frac{m_B}{m_A}\right) \sqrt{\frac{R_B}{R_A}}
\)
Substitute the given values:
\(
\frac{L_B}{L_A}=(4 \sqrt{2}) \times \sqrt{2}
\)
Simplify the expression:
\(
\begin{gathered}
\frac{L_B}{L_A}=4 \times(\sqrt{2} \times \sqrt{2}) \\
\frac{L_B}{L_A}=4 \times 2 \\
\frac{L_B}{L_A}=8
\end{gathered}
\)
The ratio of the angular momentum of planet \(B\) to that of planet \(A\) is 8.

Hint

(a) To find the new acceleration due to gravity, we need to look at the relationship between gravity, mass, and radius.
Step 1: The Formula
The acceleration due to gravity ( \(g\) ) on the surface of a planet is given by the formula:
\(
g=\frac{G M}{R^2}
\)
Where:
\(G\) is the Gravitational Constant.
\(M\) is the mass of the planet.
\(R\) is the radius of the planet.
Step 2: The Relationship
Since the mass \(M\) remains unchanged and \(G\) is a constant, we can see that:
\(
g \propto \frac{1}{R^2}
\)
Step 3: Change in Radius
The problem states that the diameter is reduced to one-third of its original value. Since \(R= \frac{D}{2}\), if the diameter is reduced to \(1 / 3\), the radius is also reduced to \(1 / 3\) :
\(
R^{\prime}=\frac{R}{3}
\)
Step 4: Calculation of New Gravity ( \(g^{\prime}\) )
Substitute the new radius into our proportionality:
\(
g^{\prime}=\frac{G M}{\left(R^{\prime}\right)^2}=\frac{G M}{(R / 3)^2}
\)
Simplify the denominator:
\(
\begin{gathered}
g^{\prime}=\frac{G M}{R^2 / 9} \\
g^{\prime}=9\left(\frac{G M}{R^2}\right) \\
g^{\prime}=9 g
\end{gathered}
\)
The acceleration due to gravity will become \(9 g\).

Hint

(c) Step 1: Determine orbital radius and speed
The satellite revolves at a height \(h=\frac{R}{3}\) from the Earth’s surface. The orbital radius \(r\) is the sum of the Earth’s radius and the height:
\(
r=R+h=R+\frac{R}{3}=\frac{4 R}{3}
\)
The orbital speed \(\boldsymbol{v}\) for a satellite in a circular orbit around Earth (mass \(\boldsymbol{M}\) ) at radius \(\boldsymbol{r}\) is given by:
\(
v=\sqrt{\frac{G M}{r}}=\sqrt{\frac{G M}{4 R / 3}}=\sqrt{\frac{3 G M}{4 R}}
\)
Step 2: Calculate the angular momentum
The angular momentum \(L\) of the satellite (mass \(m=\frac{M}{2}\) ) is given by \(L=m v r\) :
\(
\begin{gathered}
L=\left(\frac{M}{2}\right) \times\left(\sqrt{\frac{3 G M}{4 R}}\right) \times\left(\frac{4 R}{3}\right) \\
L=\frac{M \times 4 R}{2 \times 3} \times \sqrt{\frac{3 G M}{4 R}}=\frac{2 M R}{3} \sqrt{\frac{3 G M}{4 R}}
\end{gathered}
\)
Step 3: Compare with the given expression
To compare the derived \(\boldsymbol{L}\) with the given expression \(\boldsymbol{L}=\boldsymbol{M} \sqrt{\frac{\boldsymbol{G} \boldsymbol{M R}}{\boldsymbol{x}}}\), we move terms into the square root:
\(
L=\sqrt{\left(\frac{2 M R}{3}\right)^2 \times \frac{3 G M}{4 R}}
\)
\(
\begin{aligned}
L & =\sqrt{\frac{4 M^2 R^2}{9}} \times \frac{3 G M}{4 R} \\
L & =\sqrt{\frac{12 G M^3 R^2}{36 R}}=\sqrt{\frac{G M^3 R}{3}}
\end{aligned}
\)
We can factor out \(\boldsymbol{M}\) from the square root to match the format:
\(
L=\sqrt{M^2 \times \frac{G M R}{3}}=M \sqrt{\frac{G M R}{3}}
\)
Comparing \(M \sqrt{\frac{G M R}{3}}\) with \(M \sqrt{\frac{G M R}{x}}\), we find \(x=3\).

Hint

(b) Step 1: Apply Conservation of Angular Momentum
We assume the Earth is a solid sphere and its mass remains constant. When the radius changes, the angular momentum \(L\) is conserved, so the initial angular momentum \(L_1\) equals the final angular momentum \(L_2\), as no external torque acts on the system:
\(
L_1=L_2
\)
Step 2: Use Moment of Inertia and Angular Velocity Formulas
The angular momentum \(L\) is the product of moment of inertia \(I\) and angular velocity \(\omega (L=I \omega)\). For a solid sphere, \(I=\frac{2}{5} M R^2\), and \(\omega=\frac{2 \pi}{T}\), where \(T\) is the time period (duration of the day).
\(
\begin{aligned}
I_1 \omega_1 & =I_2 \omega_2 \\
\frac{2}{5} M R_1^2 \frac{2 \pi}{T_1} & =\frac{2}{5} M R_2^2 \frac{2 \pi}{T_2}
\end{aligned}
\)
Step 3: Solve for the Final Time Period ( \(\boldsymbol{T}_{\mathbf{2}}\) )
Canceling the constant terms, we get \(\frac{R_1^2}{T_1}=\frac{R_2^2}{T_2}\). Rearranging for \(T_2\) gives:
\(
T_2=T_1\left(\frac{R_2}{R_1}\right)^2
\)
Given \(T_1=24\) hours and \(R_2=\frac{3}{4} R_1\) :
\(
T_2=24 \text { hours } \times\left(\frac{3}{4}\right)^2=24 \times \frac{9}{16}=13.5 \text { hours }
\)
The duration of the day will be 13 hours and 30 minutes.

Hint

(c) Step 1: Calculate gravity at height
The acceleration due to gravity ( \(g^{\prime}\) ) at a height \(h\) from the Earth’s surface, where \(R_e\) is the radius of Earth, is given by the formula \(g^{\prime}=g_0\left(\frac{R_e}{R_e+h}\right)^2\). Given that \(h\) is equal to the radius of the earth (\(h=R_e\)) and the surface gravity is \(g_0=\pi^2 \mathrm{~m} / \mathrm{s}^{-2}\), we can calculate \(g^{\prime}\) :
\(
g^{\prime}=\pi^2\left(\frac{R_e}{R_e+R_e}\right)^2=\pi^2\left(\frac{R_e}{2 R_e}\right)^2=\pi^2\left(\frac{1}{2}\right)^2=\frac{\pi^2}{4} \mathrm{~m} / \mathrm{s}^{-2}
\)
Step 2: Calculate the time period
The time period ( \(T\) ) of a simple pendulum with length \(L\) at this location is calculated using the formula \(T=2 \pi \sqrt{\frac{L}{g^{\prime}}}\). Using the given length \(L=4 \mathrm{~m}\) and the calculated \(g^{\prime}\)
\(
T=2 \pi \sqrt{\frac{4}{\frac{\pi^2}{4}}}=2 \pi \sqrt{\frac{16}{\pi^2}}=2 \pi\left(\frac{4}{\pi}\right)=8 \mathrm{~s}
\)
The time period of small oscillations will be 8 s.

Hint

(d)

Step 1: Relate the change in volume to the change in radius
Assuming the Earth is a uniform sphere, its volume \(V\) is given by \(V=\frac{4}{3} \pi R^3\). The problem states the new volume \(V^{\prime}\) is \(\frac{1}{64}\) th of the original volume \(V\).
\(
\frac{4}{3} \pi R^3=\frac{1}{64}\left(\frac{4}{3} \pi R^3\right)
\)
Solving for the new radius \(R^{\prime}\), we find \(R^{\prime}=\frac{R}{4}\).
Step 2: Apply the conservation of angular momentum
Since no external torque acts on the Earth during this change, the angular momentum \(L\) is conserved ( \(L=L^{\prime}\) ). Angular momentum is \(L=I \omega\), where \(I\) is the moment of inertia and \(\omega\) is the angular velocity. For a uniform sphere, the moment of inertia is \(I=\frac{2}{5} M R^2\), and angular velocity is \(\omega=\frac{2 \pi}{T}\).
\(
\begin{aligned}
I \omega & =I^{\prime} \omega^{\prime} \\
\left(\frac{2}{5} M R^2\right)\left(\frac{2 \pi}{T}\right) & =\left(\frac{2}{5} M^{\prime} R^2\right)\left(\frac{2 \pi}{T^{\prime}}\right)
\end{aligned}
\)
The mass \(M\) remains the same \(\left(M^{\prime}=M\right)\). Substituting \(R^{\prime}=\frac{R}{4}\) and simplifying the equation:
\(
\begin{aligned}
\frac{R^2}{T} & =\frac{(R / 4)^2}{T^{\prime}} \\
\frac{1}{T} & =\frac{1}{16 T^{\prime}} \\
T^{\prime} & =\frac{T}{16}
\end{aligned}
\)
Step 3: Determine the value of \(\mathbf{x}\)
The original period \(T\) is 24 h. The new period is \(T^{\prime}=\frac{24}{x} \mathrm{~h}\).
\(
T^{\prime}=\frac{24 \mathrm{~h}}{16}=\frac{24}{x} \mathrm{~h}
\)
Comparing the expressions, we get \(x=16\).

Hint

(b) To solve for \(\alpha\), we need to use the approximation formulas for the acceleration due to gravity above and below the Earth’s surface, specifically for cases where the distance \(h\) is much smaller than the Earth’s radius \(R_e\).
Step 1: Gravity at a Height ( \(h \ll R_e\) )
The acceleration due to gravity \(g_h\) at a height \(h\) above the surface is given by the formula:
\(
g_h=g\left(1+\frac{h}{R_e}\right)^{-2}
\)
Using the binomial expansion for \(h \ll R_e\), this approximates to:
\(
g_h \approx g\left(1-\frac{2 h}{R_e}\right)
\)
Step 2: Gravity at a Depth (\(d\))
The acceleration due to gravity \(g_d\) at a depth \(d\) below the surface is given by:
\(
g_d=g\left(1-\frac{d}{R_e}\right)
\)
In this problem, the depth is given as \(d=\alpha h\). Substituting this:
\(
g_d=g\left(1-\frac{\alpha h}{R_e}\right)
\)
Step 3: Equating the Two Expressions
The problem states that the acceleration at height \(h\) is equal to the acceleration at depth \(\alpha h\) (\(g_h=g_d):\)
\(
g\left(1-\frac{2 h}{R_e}\right)=g\left(1-\frac{\alpha h}{R_e}\right)
\)
Solving we get, \(\alpha=2\)

Hint

(a) Step 1: Determine the orbital speed formula
The orbital speed \((v)\) of a satellite in a circular orbit around a planet with mass \(M\) and orbital radius \(R\) is given by the formula:
\(
v=\sqrt{\frac{G M}{R}}
\)
where \(\boldsymbol{G}\) is the universal gravitational constant.
Step 2: Form the ratio of speeds
The ratio of the speed of satellite \(S_1\left(v_1\right)\) to the speed of satellite \(S_2\left(v_2\right)\) is:
\(
\frac{v_1}{v_2}=\frac{\sqrt{\frac{G M}{R_1}}}{\sqrt{\frac{G M}{R_2}}}=\sqrt{\frac{R_2}{R_1}}
\)
Step 3: Substitute values and calculate the ratio
Given radii are \(R_1=3200 \mathrm{~km}\) and \(R_2=800 \mathrm{~km}\). Substituting these values into the ratio formula:
\(
\frac{v_1}{v_2}=\sqrt{\frac{800 \mathrm{~km}}{3200 \mathrm{~km}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}
\)
Step 4: Determine the value of \(x\)
The problem states that the ratio \(\frac{v_1}{v_2}\) is equal to \(\frac{1}{x}\).
\(
\frac{1}{x}=\frac{1}{2}
\)
Therefore, \(\mathrm{x}=2\).

Hint

(c)

To find the angular speed of the farther satellite as observed from the nearer one, we need to determine their relative velocity and their distance at the moment of closest approach.
Step 1: Determine the Orbital Radius of the Farther Satellite
According to Kepler’s Third Law, the square of the time period is proportional to the cube of the orbital radius (\(T^2 \propto R^3\)):
\(
\frac{R_2^3}{R_1^3}=\frac{T_2^2}{T_1^2}
\)
Given:
\(T_1=1 \mathrm{~h}, R_1=2 \times 10^3 \mathrm{~km}\)
\(T_2=8 \mathrm{~h}\)
\(
\begin{gathered}
\left(\frac{R_2}{R_1}\right)^3=\left(\frac{8}{1}\right)^2=64 \\
\frac{R_2}{R_1}=\sqrt[3]{64}=4 \\
R_2=4 \times R_1=8 \times 10^3 \mathrm{~km}
\end{gathered}
\)
Step 1: Calculate Angular and Linear Velocities
The angular velocity \(\omega\) of each satellite relative to the planet is \(\omega=\frac{2 \pi}{T}\).
For \(S_1: \omega_1=\frac{2 \pi}{1}=2 \pi \mathrm{rad} / \mathrm{h}\)
For \(S_2: \omega_2=\frac{2 \pi}{8}=\frac{\pi}{4} \mathrm{rad} / \mathrm{h}\)
The linear orbital velocity is \(v=\omega R\) :
\(v_1=(2 \pi) \times\left(2 \times 10^3\right)=4 \pi \times 10^3 \mathrm{~km} / \mathrm{h}\)
\(v_2=\left(\frac{\pi}{4}\right) \times\left(8 \times 10^3\right)=2 \pi \times 10^3 \mathrm{~km} / \mathrm{h}\)
Step 3: Relative Angular Velocity at Closest Approach
When the satellites are closest, they lie on the same radial line from the planet. Since they move in the same direction (anticlockwise), their relative velocity \(v_{\text {rel }}\) is:
\(
v_{r e l}=v_1-v_2=4 \pi \times 10^3-2 \pi \times 10^3=2 \pi \times 10^3 \mathrm{~km} / \mathrm{h}
\)
The distance between them at this instant is:
\(
r_{r e l}=R_2-R_1=8 \times 10^3-2 \times 10^3=6 \times 10^3 \mathrm{~km}
\)
The angular speed of \(S_2\) as observed from \(S_1\) is given by the formula for relative angular velocity:
\(
\begin{gathered}
\omega_{o b s}=\frac{v_{r e l}}{r_{r e l}} \\
\omega_{o b s}=\frac{2 \pi \times 10^3}{6 \times 10^3}=\frac{2 \pi}{6}=\frac{\pi}{3} \mathrm{rad} / \mathrm{h}
\end{gathered}
\)
Step 4: Solve for \({x}\)
Comparing \(\frac{\pi}{3}\) with the given form \(\frac{\pi}{x}\) :
\(
x=3
\)

Hint

(b) To find the value of \(x\) for which the gravitational potential energy of the system is maximum, we need to analyze the total potential energy based on the arrangement of the four masses at the vertices of a square.
Step 1: Identify the Potential Energy Components
Let the side length of the square be \(a\). The four masses placed at the vertices are:
\(m_1=m\)
\(m_2=M-m\)
\(m_3=m\)
\(m_4=M-m\)
In a square, there are 6 pairs of interactions:
4 Side pairs (distance \(a\) ): \(\left(m_1, m_2\right),\left(m_2, m_3\right),\left(m_3, m_4\right)\), and \(\left(m_4, m_1\right)\).
2 Diagonal pairs (distance \(\sqrt{2} a\) ): (\(m_1, m_3\) ) and ( \(m_2, m_4\)).
Step 2: Formulate the Total Potential Energy (\(U\))
The formula for gravitational potential energy is \(U=-\frac{G m_1 m_2}{r}\). The total energy \(U\) is:
\(
U=-G\left[\frac{m(M-m)+(M-m) m+m(M-m)+(M-m) m}{a}+\right.\left.\frac{m \cdot m+(M-m)(M-m)}{\sqrt{2} a}\right]
\)
Simplifying the side terms \((4 \times m(M-m))\) :
\(
U=-\frac{G}{a}\left[4 m(M-m)+\frac{m^2+(M-m)^2}{\sqrt{2}}\right]
\)
Step 3: Maximize the Energy
For \(U\) to be maximum, the term inside the brackets (let’s call it \(f(m)\)) must be minimized, because the overall expression is negative.
Note: Gravitational potential energy is always negative. A “maximum” value means the least negative value (closest to zero).
\(
\begin{gathered}
f(m)=4 m M-4 m^2+\frac{m^2+M^2-2 m M+m^2}{\sqrt{2}} \\
f(m)=4 m M-4 m^2+\frac{2 m^2-2 m M+M^2}{\sqrt{2}}
\end{gathered}
\)
To find the extremum, we differentiate \(f(m)\) with respect to \(m\) and set it to zero \(\left(\frac{d f}{d m}=0\right)\) :
\(
\frac{d f}{d m}=4 M-8 m+\frac{4 m-2 M}{\sqrt{2}}=0
\)
Multiply by \(\sqrt{2}\) to clear the fraction:
\(
\begin{gathered}
4 \sqrt{2} M-8 \sqrt{2} m+4 m-2 M=0 \\
M(4 \sqrt{2}-2)=m(8 \sqrt{2}-4)
\end{gathered}
\)
Step 4: Solve for the Ratio \(M / m\)
\(
\frac{M}{m}=\frac{8 \sqrt{2}-4}{4 \sqrt{2}-2}
\)
Factor out a 2 from the numerator:
\(
\begin{gathered}
\frac{M}{m}=\frac{2(4 \sqrt{2}-2)}{4 \sqrt{2}-2} \\
\frac{M}{m}=2
\end{gathered}
\)
The ratio \(M / m\) is \(2: 1\), so the value of \(x\) is 2.

Hint

(c)

To solve this, we need to find the “Neutral Point” between the two planets. The satellite only needs enough energy to reach this point; once it crosses it, the gravitational pull of the second planet will naturally pull it to its surface.
Step 1: Locate the Neutral Point ( \(P\) )
At the neutral point, the gravitational forces from both planets are equal and opposite. Let \(x\) be the distance of point \(P\) from the center of the first planet (mass \(M\)). The distance from the center of the second planet (mass \(9 M\)) will be \(8 R-x\).
Equating the gravitational fields:
\(
\frac{G M}{x^2}=\frac{G(9 M)}{(8 R-x)^2}
\)
Taking the square root of both sides:
\(
\begin{gathered}
\frac{1}{x}=\frac{3}{8 R-x} \\
8 R-x=3 x \Longrightarrow 4 x=8 R \Longrightarrow x=2 R
\end{gathered}
\)
The neutral point is at a distance of \(2 R\) from the center of planet 1 and \(6 R\) from the center of planet 2.
Step 2: Apply Conservation of Energy
The minimum speed \(v\) is required to take the satellite from the surface of planet \(1(r=R)\) to the neutral point \((r=2 R)\).
Initial Energy (\(E_{\hat{i}}\)):
\(
\begin{gathered}
E_i=K E+P E_1+P E_2 \\
E_i=\frac{1}{2} m v^2-\frac{G M m}{R}-\frac{9 G M m}{7 R}
\end{gathered}
\)
(Note: At the surface of planet 1 , the distance to planet 2 is \(8 R-R=7 R\))
Energy at Neutral Point (\(E_P\)): At minimum speed, the satellite’s velocity at \(P\) is essentially zero.
\(
\begin{gathered}
E_P=0-\frac{G M m}{2 R}-\frac{9 G M m}{6 R} \\
E_P=-\frac{G M m}{2 R}-\frac{3 G M m}{2 R}=-\frac{4 G M m}{2 R}=-\frac{2 G M m}{R}
\end{gathered}
\)
Step 3: Solving for \(v\)
Set \(E_i=E_P\) :
\(
\begin{gathered}
\frac{1}{2} m v^2-\frac{G M m}{R}-\frac{9 G M m}{7 R}=-\frac{2 G M m}{R} \\
\frac{1}{2} v^2=\frac{G M}{R}+\frac{9 G M}{7 R}-\frac{2 G M}{R} \\
\frac{1}{2} v^2=\frac{9 G M}{7 R}-\frac{G M}{R} \\
\frac{1}{2} v^2=\frac{9 G M-7 G M}{7 R}=\frac{2 G M}{7 R} \\
v^2=\frac{4 G M}{7 R} \Longrightarrow v=\sqrt{\frac{4}{7}} \frac{G M}{R}
\end{gathered}
\)
Step 4: Comparison
Comparing this with the given form \(v=\sqrt{\frac{a}{7} \frac{G M}{R}}\) :
\(
a=4
\)

Hint

(a) To find the new radius of the Earth after compression, we use the formula for escape velocity and examine how it scales with the radius when the mass remains constant.
Step 1: The Formula for Escape Velocity
The escape velocity ( \(v_e\) ) from the surface of a planet is given by:
\(
v_e=\sqrt{\frac{2 G M}{R}}
\)
Where:
\(G\) is the Gravitational Constant.
\(M\) is the mass of the Earth (which remains unchanged during compression).
R is the radius of the Earth.
Step 2: Establishing the Relationship
Since \(G\) and \(M\) are constant, the escape velocity is inversely proportional to the square root of the radius:
\(
v_e \propto \frac{1}{\sqrt{R}}
\)
This can be written as a ratio between the initial state (1) and the compressed state (2):
\(
\frac{v_{e 2}}{v_{e 1}}=\sqrt{\frac{R_1}{R_2}}
\)
Step 3: Step-by-Step Calculation
We are given:
Initial radius \(\left(R_1\right)=6400 \mathrm{~km}\)
The escape velocity increases 10 times, so \(\frac{v_{c 2}}{v_{e 1}}=10\)
Substitute these into the ratio:
\(
10=\sqrt{\frac{6400}{R_2}}
\)
Square both sides to remove the square root:
\(
100=\frac{6400}{R_2}
\)
Solve for \(R_2\) :
\(
\begin{aligned}
& R_2=\frac{6400}{100} \\
& R_2=64 \mathrm{~km}
\end{aligned}
\)
The radius of the Earth must be compressed to \(\mathbf{6 4 ~ k m}\).

Hint

(c) Step 1: Recall the gravitational binding energy formula
The gravitational binding energy of a uniform sphere (like the Earth, as approximated in this problem) is a standard formula that describes the total potential energy stored in the sphere due to its own gravity. The formula for this energy is:
\(
U=-\frac{3}{5} \frac{G M^2}{R}
\)
Step 2: Determine the energy required to disassemble the sphere
The energy required to completely break up the Earth and move all its mass to infinity is equal to the negative of its gravitational binding energy. This is because at infinite separation, the final potential energy is zero.
The energy \(\boldsymbol{E}\) needed is:
\(
E=0-U=-U=-\left(-\frac{3}{5} \frac{G M^2}{R}\right)=\frac{3}{5} \frac{G M^2}{R}
\)
Step 3: Compare with the given expression
The problem states that the amount of energy needed is given by the expression \(\frac{x}{5} \frac{G M^2}{R}\). Comparing this to our derived value:
\(
\frac{x}{5} \frac{G M^2}{R}=\frac{3}{5} \frac{G M^2}{R}
\)
By comparing the coefficients, we find that \(x=3\).

Note: Step-by-Step Derivation (Brief)
If you have a sphere of radius \(r\), its mass is \(m=\rho\left(\frac{4}{3} \pi r^3\right)\). If you bring an additional shell of mass \(d m=\rho\left(4 \pi r^2 d r\right)\) from infinity, the work done \(d U\) is:
\(
d U=-\frac{G m d m}{r}
\)
Substituting \(m\) and \(d m\) in terms of density \(\rho\) and integrating from \(r=0\) to \(r=R\) yields the \(\frac{3}{5}\) coefficient.

Hint

(d) Step 1: Define acceleration due to gravity formulas
We use the standard formulas for acceleration due to gravity (g) at a height \(h\) above the surface and a depth \(d\) below the surface, where \(R\) is the radius of the Earth and \(g\) is the acceleration at the surface (point B ):
At point C (height \(h\)): \(g_C=g \frac{\boldsymbol{R}^2}{(\boldsymbol{R}+\boldsymbol{h})^2}\)
At point A (depth \(d\) ): \(g_A=g\left(1-\frac{d}{R}\right)\)
Step 2: Use the given information and equate the accelerations
The problem states that \(g_A=g_C\). Context from the exam paper (implied in search results) suggests that point C is at a height \(h=R / 2=3200 \mathrm{~km}\) from the surface (or distance \(1.5 R\) from the center).
Equating \(g_A\) and \(g_C\) :
\(
g\left(1-\frac{d}{R}\right)=g \frac{R^2}{(R+h)^2}
\)
Substituting \(h=R / 2\) :
\(
\begin{gathered}
1-\frac{d}{R}=\frac{R^2}{(R+R / 2)^2} \\
1-\frac{d}{R}=\frac{R^2}{(3 R / 2)^2} \\
1-\frac{d}{R}=\frac{R^2}{9 R^2 / 4} \\
1-\frac{d}{R}=\frac{4}{9}
\end{gathered}
\)
Step 3: Solve for the depth (AB) and distance from center (OA)
From the above equation:
\(
\begin{gathered}
\frac{d}{R}=1-\frac{4}{9} \\
\frac{d}{R}=\frac{5}{9}
\end{gathered}
\)
So, the depth \(d\) (which is the distance \(A B\) ) is \(A B=\frac{5 R}{9}\).
The distance of point A from the center O is \(\boldsymbol{O A}=\boldsymbol{R}-\boldsymbol{d}\) :
\(
\begin{gathered}
O A=R-\frac{5 R}{9} \\
O A=\frac{4 R}{9}
\end{gathered}
\)
The ratio \(O A: A B\) is \(\frac{4 R}{9}: \frac{5 R}{9}\), which simplifies to \(4: 5\).
The value of \(O A: A B\) is \(4: 5\). Given the ratio is \(x: y\), the value of \(x\) is 4.

Hint

(b) To solve this, we use the Law of Conservation of Energy. For a body to reach a specific height, the total mechanical energy at the surface must equal the total mechanical energy at the maximum height.
Step 1: Energy at the Surface (Initial)
At the surface of the Earth \((r=R)\), the body has both kinetic energy and gravitational potential energy:
\(
E_i=\frac{1}{2} m v_i^2-\frac{G M m}{R}
\)
Step 2: Energy at Maximum Height (Final)
At the maximum height \(h=10 R\), the distance from the center of the Earth is \(r=R+ 10 R=11 R\). At this point, the velocity is zero:
\(
E_f=0-\frac{G M m}{11 R}
\)
Step 3: Conservation of Energy
Equating \(E_i\) and \(E_f\) :
\(
\begin{gathered}
\frac{1}{2} m v_i^2-\frac{G M m}{R}=-\frac{G M m}{11 R} \\
\frac{1}{2} v_i^2=\frac{G M}{R}-\frac{G M}{11 R} \\
\frac{1}{2} v_i^2=\frac{G M}{R}\left(1-\frac{1}{11}\right)=\frac{G M}{R}\left(\frac{10}{11}\right) \\
v_i^2=\frac{20}{11} \frac{G M}{R}
\end{gathered}
\)
Step 4: Expressing in terms of Escape Velocity
The escape velocity \(v_e\) is defined as \(v_e=\sqrt{\frac{2 G M}{R}}\), which means \(v_e^2=\frac{2 G M}{R}\). We can rewrite our expression for \(v_i^2\) to include this:
\(
\begin{gathered}
v_i^2=\frac{10}{11}\left(\frac{2 G M}{R}\right) \\
v_i^2=\frac{10}{11} v_e^2
\end{gathered}
\)
\(
v_i=\sqrt{\frac{10}{11}} v_e
\)
Step 5: Comparison
Comparing this with the given form \(v_i=\sqrt{\frac{x}{y}} v_e\) :
\(x=10\)
\(y=11\)

Hint

(d) Step 1: Apply the Law of Conservation of Energy
The total mechanical energy (kinetic plus potential) of the asteroid is conserved as it moves through the Earth’s gravitational field.
The initial total energy (\(E_i\)) at distance \(r_i=10 R\) equals the final total energy (\(E_f\)) at distance \(\boldsymbol{r}_{\boldsymbol{f}} \boldsymbol{=} \boldsymbol{R}\) (the Earth’s surface):
\(
\begin{aligned}
E_i & =E_f \\
\frac{1}{2} m v_i^2-\frac{G M m}{r_i} & =\frac{1}{2} m v_f^2-\frac{G M m}{r_f}
\end{aligned}
\)
where \(m\) is the asteroid’s mass, \(M\) is the Earth’s mass, \(G\) is the gravitational constant, \(v_i\) is the initial speed, and \(v_f\) is the final speed.
Step 2: Relate GM to Escape Velocity
The escape velocity from the Earth’s surface is given by the formula \(v_{e s c}=\sqrt{\frac{2 G M}{R}}\).
We can rearrange this to express the term \(G M\) in terms of \(v_{\text {esc }}\) and \(R\) :
\(
G M=\frac{v_{e s c}^2 R}{2}
\)
Step 3: Solve for the Final Speed
Substitute the distances \(\left(r_i=10 R, r_f=R\right)\) and the expression for \(G M\) into the energy conservation equation. After canceling the mass \(m\) of the asteroid, we get:
\(
\frac{1}{2} v_i^2-\frac{G M}{10 R}=\frac{1}{2} v_f^2-\frac{G M}{R}
\)
Rearranging to solve for \(v_f^2\) yields:
\(
\begin{gathered}
v_f^2=v_i^2+\frac{2 G M}{R}\left(1-\frac{1}{10}\right) \\
v_f^2=v_i^2+\frac{2 G M}{R}\left(\frac{9}{10}\right)
\end{gathered}
\)
Substituting \(G M=\frac{v_{\text {esc }}^2 R}{2}\) into the equation:
\(
\begin{gathered}
v_f^2=v_i^2+\frac{2}{R}\left(\frac{v_{e s c}^2 R}{2}\right)\left(\frac{9}{10}\right) \\
v_f^2=v_i^2+\frac{9 v_{e s c}^2}{10}
\end{gathered}
\)
Step 4: Calculate the Final Speed
Using the given values \(v_i=12 \mathrm{~km} / \mathrm{s}\) and \(v_{\text {esc }}=11.2 \mathrm{~km} / \mathrm{s}\) :
\(
\begin{gathered}
v_f^2=(12 \mathrm{~km} / \mathrm{s})^2+\frac{9 \times(11.2 \mathrm{~km} / \mathrm{s})^2}{10} \\
v_f^2=144(\mathrm{~km} / \mathrm{s})^2+\frac{9 \times 125.44(\mathrm{~km} / \mathrm{s})^2}{10} \\
v_f^2=144(\mathrm{~km} / \mathrm{s})^2+112.896(\mathrm{~km} / \mathrm{s})^2 \\
v_f^2=256.896(\mathrm{~km} / \mathrm{s})^2 \\
v_f=\sqrt{256.896(\mathrm{~km} / \mathrm{s})^2} \approx 16.028 \mathrm{~km} / \mathrm{s}
\end{gathered}
\)
Rounding to the nearest integer, the speed is \(16 \mathrm{~km} / \mathrm{s}\).
The speed of the asteroid when it hits the surface of the earth will be approximately 16 \(\mathrm{km} / \mathrm{s}\).

Hint

(c) Step 1: Define variables and total motion
Let \(\boldsymbol{H}\) be the total height, \(\boldsymbol{t}\) the total time to hit the ground, and \(\boldsymbol{g}\) the acceleration due to gravity. The initial velocity \(u\) is \(0 \mathrm{~m} / \mathrm{s}\) since the ball is dropped. Using the equation of motion \(s=u t+\frac{1}{2} g t^2\) for the entire 100 m fall:
\(
\begin{gathered}
100=0 \times t+\frac{1}{2} g t^2 \Rightarrow t^2=\frac{200}{g} \\
t=\sqrt{\frac{200}{g}}
\end{gathered}
\)
Step 2: Analyze motion before the last 0.5 s
The ball travels 19 m in the last \(\frac{1}{2} \mathrm{~s}\), which means it travels \(100 \mathrm{~m}-19 \mathrm{~m}=81 \mathrm{~m}\) in the time \(\left(t-\frac{1}{2}\right) \mathrm{s}\).
\(
\begin{gathered}
81=0 \times\left(t-\frac{1}{2}\right)+\frac{1}{2} g\left(t-\frac{1}{2}\right)^2 \\
81=\frac{1}{2} g\left(t-\frac{1}{2}\right)^2 \Rightarrow\left(t-\frac{1}{2}\right)^2=\frac{162}{g} \Rightarrow t-\frac{1}{2}=\sqrt{\frac{162}{g}}
\end{gathered}
\)
Step 3: Solve for the acceleration due to gravity (g)
Subtract the equation for \(\left(t-\frac{1}{2}\right)\) from the equation for \(t\) to eliminate \(t\) :
\(
\begin{gathered}
t-\left(t-\frac{1}{2}\right)=\sqrt{\frac{200}{g}}-\sqrt{\frac{162}{g}} \\
\frac{1}{2}=\frac{\sqrt{200}-\sqrt{162}}{\sqrt{g}}
\end{gathered}
\)
Simplify the square roots: \(\sqrt{200}=\sqrt{100 \times 2}=10 \sqrt{2}\) and \(\sqrt{162}=\sqrt{81 \times 2}=9 \sqrt{2}\).
\(
\begin{gathered}
\frac{1}{2}=\frac{10 \sqrt{2}-9 \sqrt{2}}{\sqrt{g}} \\
\frac{1}{2}=\frac{\sqrt{2}}{\sqrt{g}}
\end{gathered}
\)
Square both sides to solve for \(g\) :
\(
\begin{gathered}
\left(\frac{1}{2}\right)^2=\left(\frac{\sqrt{2}}{\sqrt{g}}\right)^2 \\
\frac{1}{4}=\frac{2}{g} \\
g=4 \times 2=8 \mathrm{~ms}^{-2}
\end{gathered}
\)
The acceleration due to gravity near the surface on that planet is \(8 \mathrm{~ms}^{-2}\).