8.19 NCERT Exemplar Problems

Very Short Answer Questions

Q17. Molecules in the air in the atmosphere are attracted by the gravitational force of the earth. Explain why all of them do not fall into the earth just like an apple falling from a tree.

Ans: Molecules experience the vertically downward force due to gravity just like an apple falling from a tree. Due to thermal motion, which is random, their velocity is not in the vertical direction. The downward force of gravity causes the density of air in the atmosphere close to earth higher than the density as we go up.

Q18. Give one example each of central force and noncentral force.

Ans: Central force; gravitational force of a point mass, the electrostatic force due to a point charge. Non-central force: spin-dependent nuclear forces, the magnetic force between two current-carrying loops.

Q19. Draw areal velocity versus time graph for mars.

Ans: 

Q20. What is the direction of areal velocity of the earth around the sun?

Ans: It is normal to the plane containing the earth and the sun as areal velocity
\(
\frac{\Delta \mathbf{A}}{\Delta t}=\frac{1}{2} \mathbf{r} \times \mathbf{v} \Delta t
\)

Q21. How is the gravitational force between two point masses affected when they are dipped in water keeping the separation between them the same?

Ans: It remains the same as the gravitational force is independent of the medium separating the masses.

Q22. Is it possible for a body to have inertia but no weight?

Ans: Yes, a body will always have mass but the gravitational force on it can be zero; for example, when it is kept at the centre of the earth.

Q23. We can shield a charge from electric fields by putting it inside a hollow conductor. Can we shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?

Ans: No

Q24. An astronaut inside a small spaceship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity?

Ans: Yes, if the size of the spaceship is large enough for him to detect the variation in \(g\).

Q25. The gravitational force between a hollow spherical shell (of radius \(R\) and uniform density) and a point mass is \(F\). Show the nature of \(F\) vs \(r\) graph where \(r\) is the distance of the point from the centre of the hollow spherical shell of uniform density.

Ans: 

Q26. Out of aphelion and perihelion, where is the speed of the earth more and why?

Ans: At perihelion because the earth has to cover greater linear distance to keep the areal velocity constant.

Q27: What is the angle between the equatorial plane and the orbital plane of
(a) Polar satellite?
(b) Geostationary satellite?

Ans: 

(a) \(90^{\circ}\)
(b) \(0^{\circ}\)

Short Answer Questions

Q28. Mean solar day is the time interval between two successive noon when the sun passes through the zenith point (meridian). Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian). By drawing an appropriate diagram showing the earth’s spin and orbital motion, show that the mean solar day is four minutes longer than the sidereal day. In other words, distant stars would rise 4 minutes early every successive day.
(Hint: you may assume a circular orbit for the earth).

Ans: Every day the earth advances in the orbit by approximately \(1^{\circ}\). Then, it will have to rotate by \(361^{\circ}\) (which we define as 1 day) to have sun at zenith point again. Since \(361^{\circ}\) corresponds to 24 hours; extra \(1^{\circ}\) corresponds to approximately 4 minutes [3 min 59 seconds].

Q29. Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the midpoint of the line joining their centres be in a stable equilibrium or unstable equilibrium? Give reason for your answer.

Ans: Consider moving the mass at the middle by a small amount
\(h\) to the right. Then the forces on it are: \(\frac{\mathrm{GMm}}{(R-h)^2}\) to the right and \(\frac{G M m}{(R+h)^2}\) to the left. The first is larger than the second. Hence the net force will also be towards the right. Hence the equilibrium is unstable.

Q30. Show the nature of the following graph for a satellite orbiting the earth.
(a) KE us orbital radius \(R\)
(b) PE vs orbital radius \(R\)
(c) TE us orbital radius \(R\).

Ans: 

Q31. Shown are several curves (Figure below). Explain with reason, which ones amongst them can be possible trajectories traced by a projectile (neglect air friction).

Ans: The trajectory of a particle under the gravitational force of the earth will be a conic section (for motion outside the earth) with the centre of the earth as a focus. Only (c) meets this requirement.

Q32. An object of mass \(m\) is raised from the surface of the earth to a height equal to the radius of the earth, that is, taken from a distance R to \(2 R\) from the centre of the earth. What is the gain in its potential energy?

Ans: 

\(\mathrm{mgR} / 2\)

Q33. A mass \(m\) is placed at \(\mathrm{P}\) a distance \(h\) along the normal through the centre \(O\) of a thin circular ring of mass \(M\) and radius \(r\) (Figure below). If the mass is removed further away such that OP becomes \(2 h\), by what factor the force of gravitation will decrease, if \(h=r\)?

Ans:
Only the horizontal component (1.e. along the line joining \(m\) and \(\mathrm{O}\) ) will survive. The horizontal component of the force on any point on the ring changes by a factor:
\(
\begin{aligned}
&{\left[\frac{2 r}{\left(4 r^2+r^2\right)^{3 / 2}}\right] \quad\left[\frac{\mu}{\left(r^2+r^2\right)^{3 / 2}}\right]} \\
&=\frac{4 \sqrt{2}}{5 \sqrt{5}}
\end{aligned}
\)

Long Answer Questions

Q34: A star like the sun has several bodies moving around it at different distances. Consider that all of them are moving in circular orbits. Let \(r\) be the distance of the body from the centre of the star and let its linear velocity be \(v\), angular velocity \(\omega\), kinetic energy \(K\), gravitational potential energy \(U\), total energy \(E\) and angular momentum \(l\). As the radius \(r\) of the orbit increases, determine which of the above quantitles increase and which ones decrease.

Ans: As \(r\) increases:
\(U\left(=-\frac{G M m}{r}\right)\) increases.
\(v_c\left(=\sqrt{\frac{G M}{r}}\right)\) decreases.
\(\omega\left(=\frac{v_c}{r} \times \frac{1}{r^{3 / 2}}\right)\) decreases.
\(K\) decreases because \(v\) increases.
\(E\) increases because \(|U|=2 \mathrm{~K}\) and \(U<\mathrm{O}\)
\(l\) increases because \(m v r \propto \sqrt{r}\).

Q35. Six point masses of mass \(m\) each are at the vertices of a regular hexagon of side \(l\). Calculate the force on any of the masses.

Ans: 

\(
\begin{aligned}
\mathrm{AB} &=\mathrm{C} \\
(\mathrm{AC}) &=2 \mathrm{AG}=2 \cdot l \cdot \frac{\sqrt{3}}{2}=\sqrt{3} l \\
\mathrm{AD} &=\mathrm{AH}+\mathrm{HJ}+\mathrm{JD} \\
&=\frac{l}{2}+l+\frac{l}{2} \\
&=2 l
\end{aligned}
\)
\(
A E=A C=\sqrt{3} l, \quad \mathrm{AF}=l
\)
Force along \(\mathrm{AD}\) due to \(m\) at \(\mathrm{F}\) and \(\mathrm{B}\)
\(
=G m^2\left[\frac{1}{l^2}\right] \frac{1}{2}+G m^2\left[\frac{1}{l^2}\right] \frac{1}{2}=\frac{G m^2}{l^2}
\)
Force along \(\mathrm{AD}\) due to masses at \(\mathrm{E}\) and \(\mathrm{C}\)
\(
=G m^2 \frac{1}{3 l^2} \cos (30)+\frac{G m^2}{3 l^2} \cos \left(30^{\circ}\right)
\)
\(
=\frac{G m^2}{3 l^2} \sqrt{3}=\frac{G m^2}{\sqrt{3} l^2} \text {. }
\)
Force due to mass \(M\) at D
\(
\begin{aligned}
&=\frac{G m^2}{4 l^2} . \\
&\therefore \text { Total Force }=\frac{G m^2}{l^2}\left[1+\frac{1}{\sqrt{3}}+\frac{1}{4}\right] .
\end{aligned}
\)

Q36. A satellite is to be placed in an equatorial geostationary orbit around the earth for communication.
(a) Calculate height of such a satellite.
(b) Find out the minimum number of satellites that are needed to cover entire earth so that at least one satellites is visible from any point on the equator.
\(\left[M=6 \times 10^{24} \mathrm{~kg}, R=6400 \mathrm{~km}, T=24 \mathrm{~h}, \mathrm{G}=6.67 \times 10^{-11}\right.\) SI units \(]\)

Ans:

(a)
\(r=\left(\frac{G M T^2}{4 \pi^2}\right)^{1 / 3}\)
\(\therefore h=\frac{G M T^2}{4 \pi^2}-R\)
\(=4.23 \times 10^7-6.4 \times 10^6\)
\(=3.59 \times 10^7 \mathrm{~m}\).
(b) \(\theta=\cos ^{-1} \frac{R}{R+h}\)
\(
\begin{aligned}
&=\cos ^{-1} \frac{1}{1+h / r} \\
&=\cos ^{-1} \frac{1}{1+5.61} \\
&=81^{\circ} 18^{\prime} \\
\therefore 2 \theta=& 162^{\circ} 36^{\prime} \\
\frac{360^{\circ}}{2 \theta} & \approx 2.21 ; \text { Hence minimum number }=3 .
\end{aligned}
\)

Q37. Earth’s orbit is an ellipse with eccentricity \(0.0167\). Thus, the earth’s distance from the sun and speed as it moves around the sun varies from day to day. This means that the length of the solar day is not constant throughout the year. Assume that earth’s spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain the variation of the length of the day during the year?

Ans:

Angular momentum and areal velocity are constant as the earth orbits the sun.
At perigee \(r_p^2 \omega_p=r_a^2 \omega_a\) at apogee.
If ‘ \(a\) ‘ is the semi-major axis of earth’s orbit, then \(r_p=a(1-e)\) and \(r_a=a(1+e)\)
\(
\begin{aligned}
&\therefore \frac{\omega_p}{\omega_{\mathrm{a}}}=\left(\frac{1+e}{1-e}\right)^2, \quad e=0.0167 \\
&\therefore \frac{\omega_{\mathrm{p}}}{\omega_{\mathrm{a}}}=1.0691
\end{aligned}
\)
Let \(\omega\) be angular speed which is the geometric mean of \(\omega_{\mathrm{p}}\) and \(\omega_{\mathrm{a}}\) and corresponds to mean solar day,
\(
\begin{aligned}
&\therefore\left(\frac{\omega_{\mathrm{p}}}{\omega}\right)\left(\frac{\omega}{\omega_{\mathrm{a}}}\right)=1.0691 \\
&\therefore \frac{\omega_{\mathrm{p}}}{\omega}=\frac{\omega}{\omega_{\mathrm{a}}}=1.034 .
\end{aligned}
\)
If \(\omega\) corresponds to \(1^{\circ}\) per day (mean angular speed), then
\(\omega_{\mathrm{p}}=1.034\) per day and \(\omega_{\mathrm{a}}=0.967\) per day. Since \(361^{\circ}=14 \mathrm{hrs}\) : mean solar day, we get \(361.034\) which corresponds to 24 hrs \(8.14^{\prime \prime}\) (8.1″ longer) and \(360.967^{\circ}\) corresponds to \(23 \mathrm{hrs} 59 \mathrm{~min} 52^{\prime \prime}\) (7.9″ smaller).

This does not explain the actual variation of the length of the day during the year.

Q38. A satellite is in an elliptic orbit around the earth with an aphelion of \(6 R\) and perihelion of \(2 R\) where \(R=6400 \mathrm{~km}\) is the radius of the earth. Find the eccentricity of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius \(6 R?\)
\(\left[\mathrm{G}=6.67 \times 10^{-11}\right.\) SI units and \(\left.M=6 \times 10^{24} \mathrm{~kg}\right]\)

Ans:

\(
\begin{aligned}
r_a &=a(1+e)=6 R \\
r_p &=a(1-e)=2 R \quad \Rightarrow e=\frac{1}{2}
\end{aligned}
\)
Conservation of angular momentum:
angular momentum at perigee \(=\) angular momentum at apogee
\(
\begin{aligned}
&\therefore m v_p r_p=m v_a r_a \\
&\therefore \frac{v_a}{v_p}=\frac{1}{3} .
\end{aligned}
\)
Conservation of Energy:
Energy at perigee \(=\) Energy at apogee
\(
\begin{aligned}
&\frac{1}{2} m v_p^2-\frac{G M m}{r_p}=\frac{1}{2} m v_a^2-\frac{G M m}{r_a} \\
&\therefore v_p^2\left(1-\frac{1}{9}\right)=-2 G M\left[\frac{1}{r_a}-\frac{1}{r_p}\right]=2 G M\left[\frac{1}{r_a}-\frac{1}{r_p}\right]
\end{aligned}
\)
\(
v_p=\frac{2 G M\left[\frac{1}{r_p}-\frac{1}{r_a}\right]^{1 / 2}}{\left[1-\left(v_a / v_p\right)\right]^2}=\left[\frac{\frac{2 G M}{R}\left[\frac{1}{2}-\frac{1}{6}\right]}{\left(1-\frac{1}{9}\right)}\right]^{1 / 2}
\)
\(
=\left(\frac{2 / 3}{8 / 9} \frac{G M}{R}\right)^{1 / 2}=\sqrt{\frac{3}{4} \frac{G M}{R}}=6.85 \mathrm{~km} / \mathrm{s}
\)
\(
v_p=6.85 \mathrm{~km} / \mathrm{s} \quad, v_a=2.28 \mathrm{~km} / \mathrm{s} \text {. }
\)
For \(r=6 R, v_c=\sqrt{\frac{G M}{6 R}}=3.23 \mathrm{~km} / \mathrm{s}\).
Hence to transfer to a circular orbit at apogee, we have to boost the velocity by \(\Delta=(3.23-2.28)=0.95 \mathrm{~km} / \mathrm{s}\). This can be done by suitably firing rockets from the satellite.

HC Verma Short Answer Questions

Q1. Can two particles be in equilibrium under the action of their mutual gravitational force? Can three particles be? Can one of the three particles be?

Ans: A particle will be in equilibrium when the net force acting on it is equal to zero. Two particles under the action of their mutual gravitational force will be in equilibrium when they revolve around a common point under the influence of their mutual gravitational force of attraction. In this case, the gravitational pull is used up in providing the necessary centripetal force. Hence, the net force on the particles is zero and they are in equilibrium. This is also true for a three-particle system.

Q2. Is there any meaning of “Weight of the earth” ?

Ans: Weight of a body is always because of its gravitational attraction with earth.As the law of gravitational attraction is universal so it applies to any two bodies (earth and sun as well). So we can define the weight of earth w.r.t a body of mass comparable or heavier than earth because of its gravitational attraction between earth and that body. But practically no body on earth has mass comparable to earth so the weight of earth will be a meaningless concept w.r.t earth’s frame.

Q3. If heavier bodies are attracted more strongly by the earth, why don’t they fall faster than the lighter bodies?

Ans: We know that acceleration due to a force on a body of mass in given by \(a=\frac{F}{m}\) If \(\mathrm{F}\) is the gravitational force acting on a body of mass \(m\), then \(a\) is the acceleration of a free falling body.
This force is given as \(F=G \frac{M m}{R^2}\)
Here, \(\mathrm{M}\) is the mass of the Earth; \(\mathrm{G}\) is the universal gravitational constant and \(\mathrm{R}\) is the radius of the Earth.
\(\therefore\) Acceleration due to gravity, \(a=\frac{F}{m}=\frac{G M}{R^2}\)
From the above relation, we can see that acceleration produced in a body does not depend on the mass of the body. So, acceleration due to gravity is the same for all bodies.

Q4. Can you think of two particles which do not exert gravitational force on each other?

Ans: Two particles which do not exert gravitational force on each other will be mass less particles. But every particle has even a little mass. Hence we cannot find two particles which do not exert gravitational force on each other. According to Newton”s universal law of gravitation, everybody in this universe exerts a gravitational force on each other.

Q5. The earth revolves round the sun because the sun attracts the earth. The sun also attracts the moon and this force is about twice as large as the attraction of the earth on the moon. Why does the moon not revolve round the sun? Or does it?

Ans: No, the moon does not revolve around the sun and it revolves around the earth which revolves around the sun due to gravity.
Explanation:
This is because, even though the sun is massive in size when compared to the earth and the gravitational force exerted by the sun on the moon is twice that of the earth on the moon, the distance between the objects plays a major role in the moon’s revolution. The moon is very far away from the sun and very closer to the earth, which makes the earth’s gravitational influence on the moon more powerful than the sun’s gravitational force.
Thus, the moon revolves around the earth and not the sun.

Q6. At noon, the sun and the earth pull the objects on the earth’s surface in opposite directions. At midnight, the sun and the earth pull these objects in same direction. Is the weight of an object, as measured by a spring balance on the earth’s surface, more at midnight as compared to its weight at noon?

Ans: The distance of the sun from the earth is \(\mathrm{R}=150 \times 10^6 \mathrm{~km}\) while the diameter of the earth is \(12800 \mathrm{~km}\) only. So the variance of distance \(R\) between noon and midnight is comparatively very small (about \(0.0085 \%\) ). The force on an object by the sun is inversely proportional to \(\mathrm{R}^2\), so the change in the force between noon and midnight is even smaller or say negligible. So, the weight measured by a spring balance at the earth’s surface at these two times will not differ as the spring balances are not sensitive enough to detect such negligible changes.

Q7. An apple falls from a tree. An insect in the apple finds that the earth is falling towards it with an acceleration \(g\). Who exerts the force needed to accelerate the earth with this acceleration \(g\)?

Ans: As the insect is inside the apple so it also falls with gravitational acceleration g now as the insect is taken as frame so now no need to consider the apple just assume that the insect is falling towards the earth freely. Now if the mass of an insect be \(\mathrm{m}\) and that of the earth be \(\mathrm{M}\) and \(\mathrm{r}\) be the distance between the insect and center of the earth at any instant of time. Then gravitational attraction between them i.e both attract each – other with the force \(\mathrm{F}=\mathrm{GMm} / \mathrm{r}^2\)

As the insect is falling with \(\mathrm{g}\) so the earth is appearing to coming towards the insect in the insect – frame. The force for this acceleration is exerted by the insect.

Q8. Suppose the gravitational potential due to a small system is \(k / r^2\) at a distance \(r\) from it. What will be the gravitational field? Can you think of any such system? What happens if there were negative masses?

Ans: The gravitational potential \(\mathrm{V}=\mathrm{k} / \mathrm{r}^2\).
Hence gravitational field \(E=-d V / d r=-\left(-2 \mathrm{k} / r^3\right)=2 \mathrm{k} / r^3\)
A hypothetical gravitational dipole system can be assumed like this, as in the case of an electric dipole in which the electric field is inversely proportional to the cube of the distance. It is hypothetical because in the case of gravitation negative masses do not occur. If there were negative masses the direction of the field would be reversed.

Q9. The gravitational potential energy of a two-particle system is derived in this chapter as \(U=-\frac{G m_1 m_2}{r}\). Does it follow from this equation that the potential energy for \(r=\infty\) must be zero? Can we choose the potential energy for \(r=\infty\) to be \(20 \mathrm{~J}\) and still use this formula? If no, what formula should be used to calculate the gravitational potential energy at separation \(r\)?

Ans: Yes. Since \(U\) is inversely proportional to \(r\), at \(r=\infty\) the gravitational potential energy must be zero. If we choose the potential energy at \(r=\infty\) to be \(20 \mathrm{~J}\), we can not use this formula, because putting \(r=\infty\), will not give the result \(20 \mathrm{~J}\).
The required formula will be, \(U=-\mathrm{Gm}_1 \mathrm{~m}_2 / r+20\).

Q10. The weight of an object is more at the poles than at the equator. Is it beneficial to purchase goods at the equator and sell them at the pole? Does it matter whether a spring balance is used or an equal-beam balance is used?

Ans: The weight of an object is more at the poles than that at the equator. In purchasing or selling goods, we measure the mass of the goods. The balance used to measure the mass is calibrated according to the place to give its correct reading. So, it is not beneficial to purchase goods at the equator and sell them at the poles. A beam balance measures the mass of an object, so it can be used here. For using a spring balance, we need to calibrate it according to the place to give the correct readings.

Q11. The weight of a body at the poles is greater than the weight at the equator. Is it the actual weight or the apparent weight we are talking about? Does your answer depends on whether only the earth’s rotation is taken into account or the flattening of the earth at the poles is also taken into account?

Ans: The weight of a body at the poles is greater than that at the equator. Here, we are talking about the actual weight of the body at that particular place.
Yes. If the rotation of the Earth is taken into account, then we are discussing the apparent weight of the body.

Q12. If the radius of the earth decreases by \(1 \%\) without changing its mass, will the acceleration due to gravity at the surface of the earth increase or decrease? If so, by what percent?

Ans: If we consider the Earth to be a perfect sphere, then the acceleration due to gravity at its surface is given by \(g=G \frac{M}{R^2}\)
Here, \(\mathrm{M}\) is the mass of Earth; \(\mathrm{R}\) is the radius of the Earth and \(\mathrm{G}\) is universal gravitational constant. If the radius of the earth is decreased by \(1 \%\), then the new radius becomes
\(
\begin{aligned}
&R^{\prime}=R-\frac{R}{100}=\frac{99}{100} R \\
&\Rightarrow R^{\prime}=0.99 R
\end{aligned}
\)
New acceleration due to gravity will be given by
\(
\begin{aligned}
&g^{\prime}=G \frac{M}{R^{\prime 2}}=G \frac{M}{(0.99 R)^2} \\
&\Rightarrow g^{\prime}=1.02 \times\left(G \frac{M}{R^2}\right)=1.02 g
\end{aligned}
\)
Hence, the value of the acceleration due to gravity increases when the radius is decreased. The percentage increase in the acceleration due to gravity is given by
\(
\begin{aligned}
&\frac{g^{\prime}-g}{g} \times 100 \\
&=\frac{0.02 g}{g} \times 100 \\
&=2 \%
\end{aligned}
\)

Q13. A nut becomes loose and gets detached from a satellite revolving around the earth. Will it land on the earth? If yes, where will it land? If no, how can an astronaut make it land on the earth?

Ans: If the nut is simply detached it will continue to orbit the Earth along with the satellite since it already has velocity equal to the orbiting satellite. This is how satellites are placed in orbit in the first place. They are taken to the required elevation and accelerated to attain the required speed and simply let loose.

If the nut is ejected with force against the direction of orbit the nut will revolve around the Earth in a decaying orbit and will eventually fall to the Earth. It will most probably burn on entry into the Earth’s atmosphere and will never make it to the ground.

Q14. Is it necessary for the plane of the orbit of a satellite to pass through the centre of the earth?

Ans: Yes. Because the centripetal force on the satellite is towards the center of the earth where the earth’s mass is apparently considered to be concentrated.

Q15. Consider earth satellites in circular orbits. A geostationary satellite must be at a height of about 36000 km from the earth’s surface. Will any satellite moving at this height be a geostationary satellite? Will any satellite moving at this height have a time period of 24 hours?

Ans: \(
\begin{aligned}
&T=\left(\frac{g R^2 T^2}{4 \pi^2}\right)^{\frac{1}{3}}-R \\
&T=\frac{4 \pi^2(h+R)^3}{g R^2} \\
&=\frac{4 \times 3.14^2 \times(36000+6400)^3 \times 10^9}{9.8 \times\left(6400 \times 10^3\right)^2} \\
&=24.097 \mathrm{Hr}
\end{aligned}
\)
Which implies that it is a geostationary satellite with time period \(=24 \mathrm{Hrs}\).

Q16. No part of India is situated on the equator. Is it possible to have a geostationary satellite which always remains over New Delhi?

Ans: No. All geostationary orbits are concentric with the equator of the Earth. Delhi does not lie over the equator.

Q17. As the earth rotates about its axis, a person living in his house at the equator goes in a circular orbit of radius equal to the radius of the earth. Why does he/she not feel weightless as a satellite passenger does?

Ans: To feel the weightlessness in a circular orbit around the earth, the person should orbit with such a speed so that the centrifugal force must be equal to the apparent weight of the person. For this condition \(g=v^2 / R\) at the earth’s surface.
This gives, \(\mathrm{v}=\sqrt{ } \mathrm{gR}=\sqrt{ }\left(9.8^{\star} 6400000\right)=7920 \mathrm{~m} / \mathrm{s}=7.92 \mathrm{~km} / \mathrm{s}\)
So only going in the circular orbit at the surface of the earth is not sufficient for the weightlessness but his speed also should be \(7.92 \mathrm{~km} / \mathrm{s}\) which is not possible for a person even in a vehicle or plane. Also, the resistance of air on the surface will make the vehicle burn at this speed.

Q18. Two satellites going in the equatorial plane have almost the same radii. As seen from the earth one moves from east to west and the other from west to east. Will they have the same time period as seen from the earth? If not, which one will have less time period?

Ans: As seen from the earth, the apparent time period for both the satellites will be different due to the rotation of the earth. The earth rotates from west o east. So, the satellite moving from east to west will have greater relative angular speed, hence, will have less time period.

Q19. A spacecraft consumes more fuel in going from the earth to the moon than it takes for a return trip. Comment on this statement.

Ans: Yes, a spacecraft consumes more fuel in going from the Earth to the Moon than it takes for the return trip. In going from the Earth to the Moon, the spacecraft has to overcome the gravitational pull of the earth. So, more fuel is consumed in going from the Earth to Moon. However, in the return trip, this gravitation pull helps the spacecraft to come back to Earth.

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