Gravitation The Law Of Falling(11th And 12th > Physics ) Questions and answers for exam Preparation

11th And 12th > Physics

GRAVITATION THE LAW OF FALLING MCQs

Total Questions : 30 | Page 1 of 3 pages

If the period of revolution of an artificial satellite just above the earth's surface is T and the density of earth is $ρ$ , then $ρ T^{2}$ is

A universal constant whose value is $\frac{3 π}{G}$
A universal constant whose value is $\frac{3 π}{2 G}$
Proportional to radius of earth R
Proportional to square of the radius of earth $R^{2}$ Here, G= universal gravitational constant

Discuss Question

Answer: Option A. -> A universal constant whose value is

\frac{3 π}{G}

:
A

Here, G = universal gravitation constant

Time period of a satellite above the earth's surface is :

$T^{2} = \frac{4 π^{2} R^{3}}{G M} = \frac{3 π}{G ⎛ ⎜ ⎝ \frac{M}{\frac{4}{3} π R^{3}} ⎞ ⎟ ⎠} = \frac{3 π}{G ρ}$

or $ρ T^{2} = \frac{3 π}{G}$ = a universal constant.

Question 2.

A small body starts falling on to the earth from a distance equal to the radius of the earth's orbit. How long will the body take to reach the sun? Express the time in terms of T, period of revolution of the earth round the sun

$\frac{T}{4 \sqrt{2}}$
$\frac{T}{2 \sqrt{2}}$
$\frac{T}{\sqrt{2}}$
T

Discuss Question

Answer: Option A. ->

\frac{T}{4 \sqrt{2}}

:
A

Let x be the distance moved by the particle in time t.

Then by conservation of energy $- \frac{G M m}{d} + 0 = - \frac{G M m}{d - x} + \frac{1}{2} m v^{2}$

Where M is mass of the sun and d is the initial separation.

$v = \sqrt{\frac{2 G M}{d}} \sqrt{\frac{x}{d - x}}$

$\frac{d x}{d t} = \sqrt{\frac{2 G M}{d}} \sqrt{\frac{x}{d - x}}$

$d t = \sqrt{\frac{2 G M}{d}} \int_{0}^{d} \sqrt{\frac{x}{d - x}} d x$

$t = \sqrt{\frac{d}{2 G M}} {[\sqrt{x (d - x)} - d s i n^{- 1} \sqrt{\frac{d - x}{d}}]}_{0}^{- 1} = \sqrt{\frac{d}{2 G M}} (d s i n^{- 1} 1)$

= $\sqrt{\frac{d}{2 G M}} \times d \times \frac{π}{2}$

But, $\frac{G M m_{e}}{d^{2}} = m_{e} ω^{2} d$

where $m_{e}$ = mass of the earth and $ω$ = angular speed of earth.

$∴ G M = ω^{2} d^{2}$

$t = \sqrt{\frac{d}{2 ω^{2} d^{2}}} \times d \times \frac{π}{2} = \frac{π}{2 \sqrt{2} ω} = \frac{π}{2 \sqrt{2} \frac{2 π}{T_{e}}} = \frac{T_{e}}{4 \sqrt{2}}$

Question 3.

The potential at the surface of a planet of mass M and radius R is assumed to be zero. Choose the most appropriate option

The potential at infinity is $\frac{G M}{R}$
The potential at the center of planet is $- \frac{G M}{R}$
Both (a) and (b) are correct
Both (a) and (b) are wrong

Discuss Question

Answer: Option C. -> Both (a) and (b) are correct
:
C

At all places potential will increase by $\frac{G M}{R}$

Question 4.

A particle is projected upward from the surface of earth (radius = R) with a speed equal to the orbital speed of a satellite near the earth's surface. The height to which it would rise is

$\sqrt{R}$
$\frac{R}{\sqrt{2}}$
R
2R

Discuss Question

Answer: Option C. -> R
:
C

$v = v_{0} = \frac{v_{e}}{\sqrt{2}}$ (orbital speed $v_{0}$ of a satellit near the earth's surface is equal to $\frac{1}{\sqrt{2}}$ times the escape velocity of a particle on earth's surface)

Now from conservetion of mechanical energy:

Decrease in kinetic energy = increase in potential energy

or $\frac{1}{2} m \frac{v_{e}^{2}}{2} = \frac{m g h}{(1 + \frac{h}{R})}$

or $\frac{1}{2} m (\frac{2 g R}{2}) = \frac{m g h}{(1 + \frac{h}{R})}$

or h = R

Question 5.

The earth revolves about the sun in an elliptical orbit with mean radius $9.3 \times 10^{7} m$ in a period of 1 year. Assuming that there are no outside influences

The earth's kinetic energy remains constant
The earth's angular momentum remains constant
The earth's potential energy remains constant
All are correct

Discuss Question

Answer: Option B. -> The earth's angular momentum remains constant
:
B

Kinetic and potential energies varies with position of earth w.r.t. sun. Angular momentum remains constant every where.

Question 6.

A geostationary satellite orbits around the earth in a circular orbit of radius 36,000 km. Then, the time period of a spy satellite orbiting a few hundred km above the earth's surface ( $R_{e}$ = 6400 km) will approximately be

$\frac{1}{2}$ h
1 h
2 h
4 h

Discuss Question

Answer: Option C. -> 2 h
:
C

Time period of a satellite very close to earth's surface is 84.6 min. Time period increases as the distance of the satellite from the surface of earth increase. So, time period of spy satellite orbiting a few hundred km above the earth's surface should be slightly greater than 84.6 min. Therefore, the more appropriate option is (c) or 2 h

Question 7.

Two satellites of masses $m_{1}$ and $m_{2} (m_{1} > m_{2})$ are revolving round the earth in circular orbits of radius $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. Which of the following statements is true regarding their speeds $v_{1}$ and $v_{2}$ ?

$v_{1} = v_{2}$
$v_{1} < v_{2}$
$v_{1} > v_{2}$
$\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}$

Discuss Question

Answer: Option B. ->

v_{1} < v_{2}

:
B

$v = \sqrt{\frac{G M}{r}} i f r_{1} > r_{2} t h e n v_{1} < v_{2}$
Orbit speed of satellite does not depends upon the mass of satellite

Question 8.

In a double star system of two stars of masses m and 2m, rotating about their centre of mass. Their time period of rotation about their centre of mass will be proportional to

$r^{3}$
r
$m^{\frac{1}{2}}$
$m^{\frac{- 1}{2}}$

Discuss Question

Answer: Option D. ->

m^{\frac{- 1}{2}}

:
D

$r_{2} = \frac{2 m r}{m + 2 m} = \frac{2 r}{3}$

$T_{2}^{2} = \frac{4 π^{2} r_{2}^{2}}{G M}$

$T_{2}^{2} = \frac{32 π^{2} r_{2}^{2}}{27 G M}$

$T_{2} \propto r^{\frac{3}{2}}$ ; $T_{2} \propto m^{\frac{- 1}{2}}$

Question 9.

Select the correct statement from the following

The orbital velocity of a satellite increases with the radius of the orbit
Escape velocity of a particle from the surface of the earth depends on the speed with which it is fired
The time period of a satellite does not depend on the radius of the orbit
The orbital velocity is inversely proportional to the square root of the radius of the orbit

Discuss Question

Answer: Option D. -> The orbital velocity is inversely proportional to the square root of the radius of the orbit
:
D

$v_{0} = \sqrt{\frac{G M}{r}}$

Question 10.

An earth satellite of mass m revolves in a circular orbit at a height h from the surface of the earth. R is the radius of the earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by