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Gravitation MCQs for NEET — Physics Questions with Answers

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A planet is moving in an elliptical orbit. If T, V, E, and L stand, respectively, for its kinetic energy, gravitational potential energy, total energy and angular momentum about the center of orbit, then

T is conserved

V is always positive

E is always negative

the magnitude of L is conserved but its direction changes continuously

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Explanation

The direction of angular momentum does not change. It remains same at each and every point of the path. We can find the direction of angular momentum using right hand thumb rule.

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In planetary motion the areal velocity of the position vector of a planet depends on the angular velocity ( $ω$ ) and the distance of the planet from the sun (r). The correct relation for areal velocity is:

$\frac{d A}{d t} α ω r$

$\frac{d A}{d t} α ω^{2} r$

$\frac{d A}{d t} α ω r^{2}$

$\frac{d A}{d t} α \sqrt{ω r}$

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Explanation

$(3) A = {πr}^{2} \frac{dA}{dt} = 2 πr \frac{dr}{dt} = 2 πrv \frac{dA}{dt} = 2 {πr}^{2} ω$

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If A is the areal velocity of a planet of mass M, its angular momentum is

$\frac{M}{A}$

2MA

$A^{2} M$

${AM}^{2}$

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Explanation

Angular momentum

= 2 x mass x Areal speed = 2 MA

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Two bodies of masses m and 4m are placed at a distance r. The gravitational potential at a point on the line joining them where the gravitational field is zero is

$- \frac{5 Gm}{r}$

$- \frac{6 Gm}{r}$

$- \frac{9 Gm}{r}$

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Explanation

$(3) L e t g r a v i t a t i o n a l f i e l d i s z e r o a t x d i s \tan c e f r o m m a s s m . \Rightarrow \frac{G m}{x^{2}} = \frac{4 G m}{{(r - x)}^{2}} \Rightarrow {(r - x)}^{2} = 4 x^{2} \Rightarrow r - x = 2 x \Rightarrow x = \frac{r}{3} P o t e n t i a l a t t h a t p o i n t : V = \frac{- G m}{x} - \frac{4 G m}{r - x} V = \frac{- G m}{\frac{r}{3}} - \frac{4 G m}{\frac{2 r}{3}} V = \frac{- 9 G m}{r}$

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The satellite of mass m orbiting around the earth in a circular orbit with a velocity v. The total energy will be:

$\frac{3}{4} {mv}^{2}$

$\frac{1}{2} {mv}^{2}$

$- \frac{1}{2} {mv}^{2}$

${mv}^{2}$

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Explanation

$E = - K = - \frac{m v^{2}}{2}$

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Magnitude of potential energy (U) and time period (T) of a satellite are related to each other as:

$T^{2} α \frac{1}{U^{3}}$

$T α \frac{1}{U^{3}}$

$T^{2} α U^{3}$

$T^{2} α \frac{1}{U^{2}}$

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Explanation

$|U| = \frac{G M m}{r} o r r α \frac{1}{U} T = \frac{2 πr}{V} = \frac{2 {πr}^{3 / 2}}{\sqrt{G m}} T^{2} α r^{3} α \frac{1}{U^{3}}$

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A projectile fired vertically upwards with a speed v escapes from the earth. If it is to be fired at 45 $°$ to the horizontal, what should be its speed so that it escapes from the earth?

$\frac{v}{\sqrt{2}}$

$\sqrt{2} v$

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Explanation

Escape velocity is independent of the angle of projection.

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Kepler's second law regarding constancy of areal velocity of a planet is a consequence of law of conservation of

Energy

Linear momentum

Angular momentum

Mass

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Explanation

$A n g u l a r m o m e n t u m = 2 \times m a s s \times a r e a l s p e e d$

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A body of super dense material with mass twice the mass of the earth but size very small compared to size of the earth starts from rest from h<<R above the Earth's surface. It reaches earth in time t:

$t = \sqrt{\frac{h}{g}}$

$t = \sqrt{\frac{2 h}{g}}$

$t = \sqrt{\frac{2 h}{3 g}}$

$t = \sqrt{\frac{4 h}{3 g}}$

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Explanation

$F o r c e b e t w e e n t h e b o d y a n d e a r t h = \frac{2 G M^{2}}{R^{2}} A c c e l e r a t i o n o f t h e b o d y = \frac{F_{g}}{2 M} = \frac{2 G M^{2}}{2 M R^{2}} = \frac{G M}{R^{2}} = g A c c e l e r a t i o n o f t h e e a r t h = \frac{F_{g}}{M} = \frac{2 G M^{2}}{M R^{2}} = \frac{2 G M}{R^{2}} = 2 g A c c e l e r a t i o n o f t h e b o d y w . r . t e a r t h = g - (- 2 g) = 3 g T i m e = \sqrt{\frac{2 h}{a}} = \sqrt{\frac{2 h}{3 g}}$

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A thin rod of length L is bent to form a semicircle. The mass of the rod is M. The gravitational potential at the centre of the circle is :

$- \frac{GM}{L}$

$- \frac{GM}{2 πL}$

$- \frac{πGM}{2 L}$

$- \frac{πGM}{L}$

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Explanation

$(4) As the rod is bent in the form of a semicircle, πR = L Radius of the circle, R = \frac{L}{π} V = - \frac{GM}{R} = - \frac{GM}{\frac{L}{π}} = - \frac{πGM}{L}$

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Yes. You can attempt every Gravitation question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.