# 19 - Gravitation Questions Answers

**If L is the angular momentum of a satellite revolving around earth is a circular orbit of radius r with speed v, then (i) L α v**

**(ii) L α r**

**(iii) L α √v**

**(iv) L α √r**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

L = m*v*r

**If potential energy of a body of mass m on the surface of earth is taken as zero then its potential energy at height h above the surface of earth is [ R is radius of earth and M is mass of earth]**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

[-GMm/(R+h)]-[-GMm/R] = GMm[1/R - 1/(R+h)]

solve ?

**Two points masses having m and 4m are placed at distance at r. The gravitational potential at a point, where gravitational field intensity zero is**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

field intensity at distance x from m = Gm/x^{2} - G4m/(r-x)^{2}

compare it to 0 and find x

then calculate gravitational potential at the point

**During motion of a planet from perihelion to aphelion the work done by gravitational force of sun on it is positive or negative.**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

change in K E = work done by all the forces

and **from perihelion to aphelion velocity decreases**

**so work done will be negative**

**If an object is projected vertically upwards with speed, half the escape speed of earth , then the maximum height attained by it is [R is radius of earth]**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

from ground to that point

1/2 m[√(2gR)/2]^{2 } + [-GMm/R] = 0 + [-GMm/(R+h)]

solve for h

**Two point masses having mass m and 2m are placed at distance d. The point on the line joining point masses, where gravitational field intensity is zero will be at distance**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

let the point will be at a distance of x from m then compare the gravitational field of the two masses at that point

**If earth suddenly stop rotating, then the weight of an object of mass m at equator will [ ω is angular speed of earth and radius R is its radius]**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

In rotating condition mass at equator is mg-mRω^{2}

if earth stops rotation then it will become mg so increases by mRω^{2}

**Three particles A,B and C each of mass m are lying at the corners of an equilateral triangle of side L. If the particle A is released keeping the particles B and C fixed, the magnitude of instantaneous acceleration of A is**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

force by the other two on A = Gmm/L^{2} each but the angle between the two forces is 60 degree so resultant force = √3Gmm/L^{2}

now calculate acceleration?

**A large number of identical point masses m are placed along x-axis, at x= 0,1,2,4.........The magnitude of gravitational force on mass at origin (x=0), will be**

**Asked By: SWATI KAPOOR**

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**Joshi sir comment**

**The magnitude of gravitational force on mass at origin (x=0), will be**

**Gmm/1 ^{2} + Gmm/2^{2} + Gmm/4^{2} + Gmm/8^{2 }**

**+ .........................**

**= [Gmm/1 ^{2}][1+(1/4)+(1/16)+....................]**

**= Gmm[1/(1-(1/4)]**

**solve?**