gravitation from previous years of iitjee

7/28/2019 Gravitation From Previous Years of IITJEE

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Gravitation - Previous Year IIT JEE Solved

Questions

Example Problem 1 (IIT JEE, 1981)

If the radius of the earth were to shrink by one percent, its mass remaining the same, the

acceleration due to gravity on the earths surface would:

a) Decrease

b) Remain unchanged

c) Increase

d) Be zero

Solution

The acceleration due to gravity is GM/R2

on the surface of the earth. When R decreases, the

acceleration will increase. So the answer is c).

[SolvedExample]

Example Problem 2 (IIT JEE, 1984)

The numerical value of the angular velocity of rotation of the earth should be _____ rad/s in order to

make the effective acceleration due to gravity at the equator equal to zero.

Solution

We just learnt that g = acceleration due to gravity = g R2cos

2 where is the angular velocity of

the earth and is the latitudinal angle.

At the equator the latitudinal angle is 0. Therefore for the acceleration to be zero, 0 = g - R 2(1) or

= (g/R)1/2

. The value of g is 9.8 m/s2

and the value of R is 6400000 m. Therefore, the angular

velocity of the earth should be 1.24 x 10-3

rad/s.


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GRAVITATION - Explained keeping IIT JEE and AIEEE questions in Perspective

mrbarlow.wordpress.com

You must have heard of how Isaac Newton saw an apple falling and wondered why is this apple

falling on the ground?.

He found out a lot about the gravitational force that we are going to study about in detail in this

chapter.

Law of Gravitation

According to Newtons law of gravitation, if there are two particles of mass m1 and m2 that are at a

distance r from each other then, the gravitational force of attraction between them is:

F = Gm1m2

r2


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This force acts on each particle along the line joining the two particles and it attracts the two

particles together.

In this expression, G is the gravitational constant and its value is 6.67 x 10-11

Nm2

/kg. Since the

value of G is so small, the gravitational force between ordinary masses that we encounter in day to

day life is negligible. For example the gravitational force between two 1 kg masses at a distance 1 m

apart is 6.67 x 10-11

N which is extremely less!

The gravitational force is a conservative force. When we talk about the gravitational force of the

earth we often shorten the name gravitationalforce to gravity.

The Value of Acceleration Due to Gravity

The acceleration with which the earth attracts any body near it is the acceleration due to gravity.

Take a look at the mass m in the picture that is being attracted by the earth which has a mass M.


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The acceleration due to gravity is GMm/r2

= GM/r2

m

If an object is kept on the surface of the earth, then r = R, the radius of the earth. In that case the

acceleration due to gravity is GM/R2

. The value of GM/R2

is approximately 9.8 m/s2

and is often

represented by the symbol g.

As the height h of the object above the earths surface increases, the value of r increases and the

acceleration due to gravity decreases.

At height h, acceleration due to gravity = GM/(R+h)2

= GM/R2

(1+h/R)2

= g/(1+h/R)2

.

Thus, the acceleration due to gravity for a body placed a height h above the earths surface is

g(1+h/R)-2.

If h is very less compared to R, then we can use binomial theorem and write:

Acceleration due to gravity at height = g(1-2h/R) when hR

b) F1/F2 = r22/r1

2if r1>R and r2>R

c) F1/F2 = r13/r2

3if r12>R

d) F1/F2 = r12/r2

2if r12

Solution

If r3)r.

If r If r >R, then F = GM/r2.

Let us first look at option a).

F1/F2 = r1/r2 if r12>R

Clearly, F1/F2 = r1/r2 only if r12

Option c) is wrong because there is no question of there being a

cubic term.

Option d) says that F1/F2 = r12/r2

2which cannot be true because that

will only happen when F is proportional to r2.

a) Hence, option b) is correct. F1/F2 = r22/r1

2if r1>R and r2>R. This is

true because when r > R, F is inversely proportional to r2.

[SolvedExample]

Example Problem 8 (IIT-JEE, 1993)


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A solid sphere of uniform density and radius 4 units is located with itscentre at the origin O of coordinates. Two spheres of equal radii 1

unit, with their centres at A (-2, 0, 0) and B(2, 0, 0) respectively, are

taken out of the solid leaving behind spherical cavities as shown in

the figure.

Then,

a) The gravitational field due to this object at the origin is zerob) The gravitational field at the point B (2, 0, 0) is zero

c) The gravitational potential is the same at all points of the circle

y2

+ z2

= 36

d) The gravitational potential is the same at all points on the circle

y2

+ z2

= 4

Solution

Whenever, a hole inside a rigid body is given, you can consider the

mass of the removed part as negative mass. This will help you

solve the problem easily.

In this case, let us look at the net electric field vectors at O because

of the large sphere and the two negative mass spheres.


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If the mass of the removed portion is M, then the gravitational field

due to each removed sphere is GM/(22) towards the centre of the

sphere. These fields cancel each other. The fields are towards the

centre of the sphere and not towards the centres of the removed

spheres because the masses of the removes spheres are negative.

The field due to the big sphere at its centre is 0 as GMr/R3

is 0 when r

= 0.

Therefore, option a) is correct.

Options c) and d) are also correct because both y2

+ z2

= 36 and y2

+

z2

= 4 represent circles that lie on the y-z plane which have the origin

at the center. So every point on each circle will be equidistant from A

and B and O. So the gravitational potential because of the removedspheres and the big sphere will be the same for all points on both

these circles.

Hence, options a), c) and d) are correct. The gravitational potential

at B will not be zero because of the removed sphere with centre A

and the big solid sphere, as well as the removed sphere with centre

B. So b) is wrong.


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[SolvedExample]


The masses and radii of the Earth and the Moon are M1, R1 and M2,

R2 respectively. Their centres are a distance d apart. The minimum

speed with which a particle of mass m should be projected from a

point midway between the two centres so as to escape to infinity is

_______.

Solution

The initial energy of the particle is:

mv2G(M1m)/(d/2) GM2m/(d/2).

The final energy is zero, because we want the particle to escape toinfinity.

Therefore, mv2G(M1m)/(d/2) GM2m/(d/2) = 0.

This gives the value of v as


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[SolvedExample]


There is a crater of depth R/100 on the surface of the moon (radius

R). A projectile is fired vertically upward from the crater with

velocity, which is equal to the escape velocity v from the surface of

the moon. Find the maximum height attained by the projectile.

Solution

We had seen that escape velocity

Conserving energy between the maximum height reached by the

projectile and the point where it starts:

m(2GM/R) (GM/R3)(1.5R2 0.5(R R/100)2]= 0 GMm/h

(Kinetic energy initially + potential energy initially = kinetic energy

finally which is zero + potential energy finally)

This gives the value of h as 101.5R.

Therefore, the height reached from the surface of the moon is 101.5

R = 99.5 R.


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Satellites

We know that the moon is the earths satellite. Similarly, there are

many artificial satellites the circle the earth. Basically, satellites are

bodies that revolve around planets like our planet earth.

Though the path of most satellites is elliptical, in most cases, we can

assume that the orbits of satellites are circular.

Take a look at a satellite revolving around the earth in a perfectly

circular orbit of radius r below.

Net force on the satellite towards the centre = centripetal forcerequired to make the satellite revolve in a circular orbit.


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The net force towards the centre is clearly GMm/r2

= mv2/r

Here r is the radius of revolution of the satellite, R is the radius of the

earth and M is the mass of the earth.

Thus, v = (GM/r)1/2

So, if a satellite revolves around a planet such that its radius of

revolution is r, then the velocity with which is should revolve is

(GM/r)1/2

.

The time period of revolution in such a case is distance travelled in

one revolution/velocity = 2r/v = 2(r3/gR

2)

1/2.

An interesting fact here is that if a satellite revolves in the same

sense as the earth is revolving with the same time period as the of

the earths rotation, then it will always e above a certain point on the

earths surface!

Such satellites which are always stationary with respect to a point on

the earths surface are called geo-stationary satellites.

Keplers Laws

In reality, planets do not move around the sun in circular orbits


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they move in elliptical orbits. The diagram below shows the motion

of the earth around the sun.

The point S represents the sun and the point E represents the earth.

The longer axis of the ellipse is called the major axis and the smaller

axis is the minor axis. Half of the length of the major axis is generally

called semi-major axis and it is represented by the symbol a. The

distance between the sun and the centre of the ellipse is ae where e

is the eccentricity of the ellipse.

The earth is closest to the sun when it is on the left end of the ellipse

the position of closest approach is called perihelion. Aphelion

represents the point when the earth at the maximum possible

distance from the sun at the right end of the ellipse.


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As the earth or any other planet moves along its elliptical path, the

area swept by the radius vector from the sun to the planet in unit

time is constant. If this area is A, we can say that dA/dt is constant.

In the figure above, if E and E represent the position of the earth at

times t and t+dt, then the area of the triangle SEE is dA. dA/dt = rate

of sweeping of the area by the radius vector joining the earth andthe sun and this is constant.

We do not use this law of Keplers directly. We generally use it

another form.

To derive this form, consider the triangle SEE below again.


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dA = area of triangle SEE = (1/2)(base)(height) = (1/2)(SE)(XE) =(1/2)(r)(EEsin)

= (1/2)(r)vdtsin) because EE = distance covered by the earth in time

dt = vdt.

Thus, dA/dt = (1/2)rvsin = constant.

Note the angular momentum of the earth about the sun. It

is or mvrsin in the direction perpendicular to the plane of

paper.

But dA/dt = (1/2)rvsin = L/2m.

Since dA/dt = constant, we can therefore say that the angular

momentum of the earth about the sun is constant.

This is the practical use of Keplers second law in our IITJEE problems.

In most problems related to Keplers laws, you will have to conserve

the angular momentum and energy between two points.

Note however that this angular momentum must be calculated

about the sun. Also note that in the expression L = mvrsin,


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represents the angle between the velocity vector of the earth and

the radius vector from the earth to the sun.

Keplers third law can be mathematically stated by the expression:

In this expression M represents the mass of the sun.

[SolvedExample]


If the distance between the earth and the sun were half its present

value, the number of days in a year would have been:

a) 64.5

b) 129c) 182.5

d) 730

Solution

In this question, since no other parameters are mentioned, we can

assume that the earth is moving around the sun in a circular orbit.


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In that case, T a3/2

.

In the case of circular motion, T r3/2

.

Therefore, T1/T2 = (r1/r2)3/2

(365/T) = (2/1)3/2

T = 129 days.

So an year would last 129 days if the distance between the earth and

sun were halved.

The answer is b).

[SolvedExample]


A satellite S is moving in an elliptical orbit around the earth. The

mass of the satellite is very small compared to the mass of the earth:

a) The acceleration of S is always directed towards the centre of

the earth

b) The angular momentum of S about the centre of the earth

changes in direction, but its magnitude remains constant

c) The total mechanical energy of S varies periodically with time

d) The linear momentum of S remains constant in magnitude

Solution


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What applies force on the satellite? The earth does and the force

applied by the earth is towards its centre. Therefore option a) iscorrect.

If you curl your fingers from the direction of the thumb

always points in the same direction perpendicular to the plane of

paper. So the angular momentum does not change its direction.

Option b) is wrong.

The force applied by the earth on the satellite is conservative, so the

net mechanical energy of the satellite is constant. Option c) is wrong.

The speed of the satellite changes as its distance from the earth

changes because the potential energy changes as the distance

changes (proportional to 1/r) and that changes the kinetic energy.

Hence the linear momentums magnitude is also not constant.

Option d) is wrong.

[SolvedExample]

Example Problem 13

A satellite moves around the earth in an elliptical orbit. Its minimum

distance from the sun is 4R and its maximum distance from the earth

is 10R. What is its maximum speed?


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Solution

The speed of the satellite will be maximum when the kinetic energy

of the satellite is maximum. The kinetic energy will be maximum

when the potential energy will be minimum or the distance of the

planet from the satellite is minimum.

Let us conserve the angular momentum and energy of the satellite at

the instant when it is at its nearest distance from the earth and whenit is farthest from the earth.

mv1(4R) = mv2(10R)

Here v1 is the velocity of the satellite when it is at its nearest distance

to the earth and v2 is the velocity of the satellite when it is farthest

from the earth. We want to find the value of v1.

Similarly, by conservation of energy:

mv12 GMm/4R = mv2

2 GMm/10R

gravitation from previous years of iitjee

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