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]
Example Problem 9 (IIT-JEE, 1988)
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]
Example Problem 10 (IIT-JEE, 2003)
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]
Example Problem 11 (IIT-JEE, 1996)
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]
Example Problem 12 (IIT-JEE, 1998)
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
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