Physics - Gravitation - Quiz-5

A satellite whose mass is m, is revolving in a circular orbit of radius r, around the earth of mass M. Time of revolution of the satellite is:

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$T \propto \frac{r^{5}}{GM}$
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$T \propto \sqrt{\frac{r^{3}}{GM}}$
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$T \propto \sqrt{\frac{r}{{GM}^{2} / 3}}$
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$T \propto \sqrt{\frac{r^{3}}{{GM}^{2} / 4}}$

Suppose the gravitational force varies inversely as the $n^{th}$ power of distance then the time period of a planet in circular orbit of radius R around the sun will be proportional to:

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$R^{(\frac{n + 1}{2})}$
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$R^{(\frac{n - 1}{2})}$
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$R^{n}$
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$R^{(\frac{n - 2}{2})}$

The distance of a geostationary satellite from the centre of the earth (Radius R = 6400 km) is nearest to:

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5R
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7R
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10R
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18R

In order to make the effective acceleration due to gravity equal to zero at the equator, the angular velocity of rotation of the earth about its axis should be (g=10 ${ms}^{- 2}$ and radius of earth is 6400 kms)

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0 rad/sec
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$\frac{1}{800}$ rad/sec
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$\frac{1}{80}$ rad/sec
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$\frac{1}{8}$ rad/sec

A body of mass m is taken from the earth surface to the height h equal to the radius of the earth, the increase in potential energy will be:

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() mgR
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() $\frac{1}{2}$ mgR
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() 2 mgR
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() $\frac{1}{4}$ mgR

Time period of a satellite revolving above Earth’s surface at a height equal to R (the radius of Earth) will be:
[g is the acceleration due to gravity at Earth’s surface]

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$2 π \sqrt{\frac{2 R}{g}}$
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$4 \sqrt{2} π \sqrt{\frac{R}{g}}$
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$2 π \sqrt{\frac{R}{g}}$
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$8 π \sqrt{\frac{R}{g}}$

An artificial satellite moving in a circular orbit around the earth has a total (kinetic + potential) energy $E_{0}$ . Its potential energy is:

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$- E_{0}$
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$1.5 E_{0}$
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$2 E_{0}$
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$E_{0}$

Given the radius of Earth ‘R’ and length of a day ‘T’, the height of a geostationary satellite is:

[G–Gravitational Constant, M–Mass of Earth]

(a) ${(\frac{4 π^{2} GM}{T^{2}})}^{\frac{1}{3}}$ (b) ${(\frac{4 πGM}{R^{2}})}^{\frac{1}{3}} - R$

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1
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2
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3
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4

A rocket of mass M is launched vertically from the surface of the earth with an initial speed v. Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is

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$R / (\frac{gR}{2 v^{2}} - 1)$
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$R (\frac{gR}{2 v^{2}} - 1)$
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$R / (\frac{2 gR}{v^{2}} - 1)$
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$R (\frac{2 gR}{v^{2}} - 1)$

Two bodies of masses $m_{1}$ and $m_{2}$ are initially at rest at infinite distance apart. They are then allowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach at a separation distance r between them is

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${[2 G \frac{(m_{1} - m_{2})}{r}]}^{1 / 2}$
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${[\frac{2 G}{r} (m_{1} + m_{2})]}^{1 / 2}$
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${[\frac{r}{2 G (m_{1} m_{2)}}]}^{1 / 2}$
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${[\frac{2 G}{r} m_{1} m_{2}]}^{1 / 2}$

The rotation period of an earth satellite close to the surface of the earth is 83 minutes. The time period of another earth satellite in an orbit at a distance of three earth radii from its surface will be

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83 minutes
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$83 \times \sqrt{8}$ minutes
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664 minutes
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249 minutes

A satellite of mass m is circulating around the earth with constant angular velocity. If the radius of the orbit is $R_{0}$ and mass of the earth M, the angular momentum about the centre of the earth is:

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$m \sqrt{G M R_{0}}$
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$M \sqrt{G m R_{0}}$
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$m \sqrt{\frac{G M}{R_{0}}}$
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$M \sqrt{\frac{G M}{R_{0}}}$

If the distance between the earth and the sun becomes half its present value, the number of days in a year would have been

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64.5
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129
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182.5
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730

A geostationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a satellite orbiting a few hundred kilometres above the earth’s surface ( $R_{earth}$ = 6400 km) will approximately be -

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1/2 h
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1 h
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2 h
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4 h

A planet revolves around the sun whose mean distance is 1.588 times the mean distance between earth and the sun. The revolution time of the planet will be:

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(1) 25 years
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(2) 1.59 years
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(3) 0.89 years
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(4) 2 years

The earth E moves in an elliptical orbit with the sun S at one of the foci as shown in figure. Its speed of motion will be maximum at the point

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C
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A
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B
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D

The period of revolution of planet A around the sun is 8 times that of B. The distance of A from the sun is how many times greater than that of B from the sun ?

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2
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3
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4
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5

If the radius of earth's orbit is made 1/4 th, the duration of an year will become -

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8 times
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4 times
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1/8 times
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1/4 times

If the mass of a satellite is doubled and the time period remain constant, the ratio of orbit in the two cases will be

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1 : 2
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1 : 1
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1 : 3
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None of these

The maximum and minimum distances of a comet from the sun are $8 \times 10^{12}$ m and $1.6 \times 10^{12}$ m. If its velocity when nearest to the sun is 60 m/s, what will be its velocity in m/s when it is farthest -

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12
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60
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112
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6

A body revolved around the sun 27 times faster than the earth. What is the ratio of their radii?

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1/3
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1/9
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1/27
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1/4

The period of the moon’s rotation around the earth is nearly 29 days. If the moon’s mass were 2 fold, its present value and all other things remained unchanged, the period of the moon’s rotation would be nearly:

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$29 \sqrt{2} d a y s$
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$29 / \sqrt{2} d a y s$
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$29 \times 2 d a y s$
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29 days

If two planets are at mean distances $d_{1}$ and $d_{2}$ from the sun and their frequencies are n₁ and n₂ respectively, then:

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${n_{1}}^{2} {d_{1}}^{2} = n_{2} {d_{2}}^{2}$
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${n_{2}}^{2} {d_{2}}^{3} = {n_{1}}^{2} {d_{1}}^{3}$
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$n_{1} {d_{1}}^{2} = n_{2} {d_{2}}^{2}$
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${n_{1}}^{2} d_{1} = {n_{2}}^{2} d_{2}$

Escape velocity of a body of 1 kg mass on a planet is 100 m/sec. Gravitational Potential energy of the body at the planet is -

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– 5000 J
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– 1000 J
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– 2400 J
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5000 J

Suppose the law of gravitational attraction suddenly changes and becomes an inverse cube law i.e., $F \propto 1 / r^{3}$ but the force still remains a central force, then:

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Kepler's law of areas still holds
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Kepler's law of period still holds
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Kepler's law of areas and period still hold
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Neither the law of areas nor the law of period still hold

A planet is revolving around the sun as shown in an elliptical path.

The correct option is:

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The time taken in travelling DAB is less than that for BCD
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The time taken in travelling DAB is greater than that for BCD
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The time taken in travelling CDA is less than that for ABC
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The time taken in travelling CDA is greater than that for ABC

In an elliptical orbit under gravitational force, in general

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Tangential velocity is constant
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Angular velocity is constant
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Radial velocity is constant
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Areal velocity is constant

Earth is revolving around the sun, if the distance of the Earth from the Sun is reduced to 1/4^th of the present distance then the present year length reduced to -

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$\frac{1}{4}$
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$\frac{1}{2}$
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$\frac{1}{8}$
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$\frac{1}{6}$

Two satellite are revolving around the earth with velocities $v_{1}$ and $v_{2}$ and in radii $r_{1}$ and $r_{2} (r_{1} > r_{2})$ respectively. Then -

(1) $v_{1} = v_{2}$

(2) $v_{1} > v_{2}$

(3) $v_{1} < v_{2}$

(4) $\frac{v_{1}}{r_{1}} = \frac{v_{2}}{r_{2}}$

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1
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2
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3
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4

If orbital velocity of planet is given by $v = G^{a} M^{b} R^{c}$ , then -

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$a = 1 / 3, b = 1 / 3, c = ‐ 1 / 3$
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$a = 1 / 2, b = 1 / 2, c = ‐ 1 / 2$
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$a = 1 / 2, b = ‐ 1 / 2, c = 1 / 2$
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$a = 1 / 2, b = ‐ 1 / 2, c = ‐ 1 / 2$

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Practice Physics Quiz Questions and Answers

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Practice Physics Quiz Questions and Answers

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