CBSE-XI-Physics
11: Gravitation
Note: Please signup/signin free to get personalized experience.
Note: Please signup/signin free to get personalized experience.
10 minutes can boost your percentage by 10%
Note: Please signup/signin free to get personalized experience.
- #Section : i
- Qstn #1Can 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.
Page No 223:
- Qstn #2Is there any meaning of “Weight of the earth”?Ans : Weight of a body is always because of its gravitational attraction with earth.As 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 weight of earth will be a meaningless concept w.r.t earth frame.
Page No 223:
- Qstn #3If 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 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{Mm}{{R}^{2}}``.
Here, M is the mass of the Earth; G is the universal gravitational constant and R is the radius of the Earth.
∴ Acceleration due to gravity, `` a=\frac{F}{m}=\frac{GM}{{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.
Page No 223:
- Qstn #4Can you think of two particles which do not exert gravitational force on each other?Ans : No. All practicals which have mass exert gravitational force on each other. Even massless particles experience the same gravitational force like other particles, because they do have relativistic mass.
Page No 223:
- Qstn #5The 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 : We know that the Earth-Moon system revolves around the Sun. The gravitational force of the Sun on the system provides the centripetal force its revolution. Therefore, the net force on the system is zero and the Moon does not experience any force from the Sun. This is the reason why the Moon revolves around the Earth and not around the Sun.
Page No 223:
- Qstn #6At 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 : No. Due to the revolution of the Earth around the Sun, the gravitational force of the Sun on the Earth system is almost zero. Hence, the body will not experience any force due to the Sun. Therefore, weight of the object will remain the same.
Page No 223:
- Qstn #7An 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 : The mutual gravitational force between the apple and the Earth is responsible for the acceleration produced in the apple falling from the tree. Although the Earth will experience the same force, it does not get attracted towards the apple because of its large mass. The insect feels that the Earth is falling towards the apple with an acceleration g because of the the relative motion.
let
vae=velocity of apple w.r.t earth
`` {\text{V}}_{ea}=velocityofearthw.r.tapple``
`` {\text{v}}_{ae}={v}_{a}-{v}_{e}=-({v}_{e}-{v}_{a})=-{v}_{ea}``
As the insect is in the frame of apple so he sees the earth moving with a relative velocity `` {\text{v}}_{ea}``.
Any other observer on earth will see the apple moving towards earth with velocity `` {\text{v}}_{ae}``.Both are opposite in direction.
Page No 223:
- Qstn #8Suppose the gravitational potential due to a small system is k/r2 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 due to the system is given as `` V=\frac{k}{{r}^{2}}``.
Gravitational field due to the system:
`` E=-\frac{dV}{dr}``
`` \Rightarrow E=-\frac{d}{dr}\left(\frac{k}{{r}^{2}}\right)=-\left(-\frac{2k}{{r}^{3}}\right)``
`` \Rightarrow E=\frac{2k}{{r}^{3}}``
We can see that for this system, `` E\propto \frac{1}{{r}^{3}}``
This type of system is not possible because `` {\text{F}}_{g}`` is always proportional to inverse of square of distance(experimental fact).
If there were negative masses, then this type of system is possible.
This system is a dipole of two masses, i.e., two masses, one positive and the other negative, separated by a small distance.
In this case, the gradational field due to the dipole is proportional to `` \frac{1}{{r}^{3}}``.
Page No 223:
- Qstn #9The gravitational potential energy of a two-particle system is derived in this chapter as
U=Gm1m2r. Does it follow from this equation that the potential energy for
r=∞must be zero? Can we choose the potential energy for
r=∞to be 20 J and still use this formula? If no, what formula should be used to calculate the gravitational potential energy at separation r?Ans : The gravitational potential energy of a two-particle system is given by `` U=-\frac{G{m}_{1}{m}_{2}}{r}``.
This relation does not tell that the gravitational potential energy is zero at infinity. For our convenience, we choose the potential energies of the two particles to be zero when the separation between them is infinity.
No, if we suppose that the potential energy for `` r=\infty `` is 20 J, then we need to modify the formula.
Now, potential energy of the two-particle system separated by a distance r is given by
`` U\left({r}^{,}\right)=U\left(r\right)-U(\infty )``
`` \,\mathrm{\,Given\,}:U(\infty )=20\,\mathrm{\,J\,}``
`` \therefore U\mathit{\left(}r\mathit{\right)}\mathit{=}\mathit{-}G\frac{{m}_{\mathit{1}}{m}_{\mathit{2}}}{{\,\mathrm{\,r\,}}^{2}}-20``
This formula should be used to calculate the gravitational potential energy at separation r.
Page No 223:
- Qstn #10The weight of an object is more at the poles than at the equator. Is it beneficial to purchase goods at 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.
Page No 223:
- Qstn #11The 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 depend 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.
Page No 224:
- Qstn #12If 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 per cent?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, M is the mass of Earth; R is the radius of the Earth and G is universal gravitational constant.
If the radius of the earth is decreased by 1%, then the new radius becomes
`` R\text{'}=R-\frac{R}{100}=\frac{99}{100}R``
`` \Rightarrow R\text{'}=0.99R``
New acceleration due to gravity will be given by
`` g\text{'}=G\frac{M}{R{\text{'}}^{2}}=G\frac{M}{(0.99R{)}^{2}}``
`` \Rightarrow g\text{'}=1.02\times \left(G\frac{M}{{R}^{2}}\right)=1.02g``
Hence, the value of the acceleration due to gravity increases when the radius is decreased.
Percentage increase in the acceleration due to gravity is given by
`` \frac{g\text{'}-g}{g}\times 100``
`` =\frac{0.02g}{g}\times 100``
`` =2\%``
Page No 224:
- Qstn #13A 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 : No, it will not land on the Earth. The nut will start revolving in the orbit of the satellite with the same orbital speed as that of the satellite due to inertia of motion. An astronaut can make it land on the Earth by projecting it with some velocity toward the Earth.
Page No 224: