Newton’s Law of Gravitation:

Every particle in the universe attracts every other particle with a force proportional to the product
of their masses and inversely proportional to the square of the distance between them. The
direction of the force is along the line joining the particles.

Thus the magnitude of the gravitational force F that two particles of masses m1 and m2 separated
by a distance r exert on each other is

G is called the gravitational constant

G= 6.673 ⅹ 10-11 Nm2/kg2
The gravitational forces between two particles are an action reaction pair. The first particle exerts
a force on the second particle that is directed toward the first particle along the line joining them.
The second particle exerts a force on the first particle that is directed toward the second particle
along the line joining them.
F21 m2 m2 m2
r21
F21

r12
m1 F12 F12
m1 m1
The force F21 exerted on m2 by m1 is directed opposite to the displacement r12 of m2 from m1.
Similarly, F12 is directed opposite to r21. F12=-F21 action reaction pair. The law of gravitation is a
vector law. The gravitational force F21 exerted on m2 by m1 is given in direction and magnitude
by the vector relation

Similarly,

The minus sign of the equation tells that F21 points in a direction opposite to r12

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Gravitational Potential Energy

From work-energy theorem we know that,

W=ΔK

Which states that the work done, W is equal to the change in kinetic energy. Conservation of
energy tells us

ΔK+ΔU=0

In which ΔU is the change in potential energy.

Conservation of energy holds for conservative force which is defined as the work done by this
force does not depend on the path but only on the initial and final position.

Where a denotes the initial point and b denotes the final point. In the figure a particle M exerts a
gravitational force F on a particle of mass m located at r. The particle of mass m is displaced a
small distance d, which is in opposite direction of r. so ds= ‒dr. m

The work done by F when the particle moves from a to b is ds F

r

M
We choose our reference configuration to be at an infinite separation of particle (ra→∞) and we
define U (∞) to be zero. At an arbitrary separation r, the potential energy is

The minus indicates that the potential energy is negative at any finite distance, that is the
potential energy is zero at infinity and decreases as the separation distance decreases. This
corresponds to the fact that the gravitational force exerted on m by M is attractive.
We can reverse the previous calculation and derive the gravitational force from the potential
energy. For spherically symmetric potential energy functions , the relation , gives the
radial component of the force with the potential energy,
The gravitational field and the gravitational potential
A basic fact of gravitation is that two particles exert forces on each other even though they are
not contact, this point of view is called action-at-a-distance. Another point of view is called

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concept which regards a particle as modifying the space around it in some way and setting up a
gravitational field. This field, the strength of which depends on the mass of the particle, then acts
on any other particle, exerting the force of gravitational attraction on it.

We define the gravitational field strength at a point as the gravitational force per unit mass at that
point, or, in terms of our mass

By moving the test mass to various positions, we can make a map showing the gravitational field
at any point in space. The gravitational field is an example of vector field, each point in this field
having a vector associated with it. The gravitational field arising from a fixed distribution of
matter is an example of a static field, because the value of the field at a given point does not
change with time.

We can also describe the gravitational field of a body by a scaler function called potential. Let us
move a test particle from infinity (where the field is zero) toward a body of mass M at a distance
r, where potential energy is U(r). We define the gravitational potential V at that point as

V(r)= U(r)/mₒ

The potential is the same as the potential energy per unit test mass. Again U(r)= ‒GMmₒ/r

The gravitational potential can be written as V(r)= ‒ GM/r. The potential V(r) is independent of
the value of the test mass mₒ. Similarly, the gravitational field g is also independent of mₒ.

Gravitational effect of a spherical distribution of mass

Let us consider a uniformly dense spherical shell of mass “M” and whose thickness “t” is small
compared to its radius “R”. we will find the gravitational force it exerts on an external particle P
of mass “m”. A small part of the shell at A attracts m with a force FA and a small part of the
shell at b, equally far from m but diametrically opposite A, attracts m with a force FB.

The resultant of these two forces on m is

FA +FB. Each of these forces has a
component Fcos along the symmetry
axis and a component F sin
perpendicular to the axis. The
perpendicular component of the Forces
cancels each other. Let us take as our
element of mass of the shell a circular
strip dM. Its radius is R sinθ, its length is
2π(Rsinθ), its width is Rdθ., its thichness
is t. Hence its volume is:
dV=2πtR2sinθdθ……….(1)
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If the density of the strip is ρ then the mass within the strip is :

dM= ρdV=2πtρR2sinθdθ……….(2)

Every particle of the ring, such as one of the mass dmA at A, attracts P with a force that has an
axial component
…….(3)
Adding the contributions for all the particles in the ring gives

or,
……………….(4)
Where, dM is the total mass of the ring and dF is the total force on m exerted by the ring.
putting the value of dM in eq n(4) we get:
……(5)

The variables x, and θ are related as

r=x cos + R cosθ
……..(6)
using the law of cosine, x2=r2+R2-2rRcosθ
now, …………(7)
putting eqn(7) & (8) in eqn (5) we get-

or,
This is the force exerted by the circular strip dM on the particle P of mass m.
Now we consider every element of mass in the shell by summing over all the circular strips in
the entire shell.
The needed integral is

so,
, where M= 4πR2tρ

A uniformly dense spherical shell attracts an external point mass as if all the mass of the shell
were concentrated at its center .

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 The force exerted by a spherical shell on a particle inside it is zero. In that case the limit
of needed integration over x is R-r to r+R, gives

So, F=0
A uniform spherical shell of matter exerts no gravitational force on a particle located inside it.

Gravitation near the earth surface

Let us assume the earth is spherical and that its density depends only on the radial distance from
its center, the magnitudes of the gravitational force acting on aparticle of mass m located at an
external point distance r from the earth’s center, can be written as

ME is the mass of the Earth. This gravitational force can also be written using Newton’s second
law of motion as
F=mg
Here, g is the free-fall acceleration due to the gravitational pull of the Earth. Combining the two
equations above gives
,
In the case of earth surface, this can be written as
,
In which gs and RE are the acceleration due to gravity at the earth surface and the radius of the
earth respectively.

Kepler’s laws of motions of motions of planets and satellites

The empirical basis for understanding the motions of the planets is Kepler’s laws, these are:
1. The laws of orbits: All plants move in elliptical orbits having the Sun at one focus.

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Newton was the first to realize that there is a direct mathematical relationship between inverse-
square (1/r2) forces and elliptical orbit. The origin of coordinates is at the central body, the
orbiting body is located at polar co-ordinates r and θ. The orbit is described by two parameters:
semi major axis a and the eccentricity e. the distance from the center of the ellipse to the either
focus is ea. A circular orbit is special case of an elliptical orbit with e=0, in which case the two
foci merge to a single point at the center of the circle. For all planets in solar system, the
eccentricities are small and the orbits are nearly circular.
2. The law of Areas: A line joining any planet to the sun Sweeps out equal areas in equal
times.

Figure ‘a’ illustrates this law; in effect it says that the orbiting body moves more rapidly when it
is closed to the central body than it does when it is far away.

Consider a small area increment ΔA covered in a time interval Δt, as shown in figure ‘b’. The
area of this approximately triangular wedge is one-half its base, r Δθ, times its height r. The rate
at which this area is swept out is ΔA/Δt= ½ (r Δθ) (r)/Δt. In the instantaneous limit this becomes

The instantaneous angular momentum of the orbiting body is L=mr2ω and so

To the extent that we can regard the two bodies as an isolated system, L is constant and therefore
dA/dt is a constant. The speeding up of a comet as it passes close to the sun is therefore just a
demonstration of the conservation of the angular momentum.

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 3. The Law of Periods: The square of the period of any planet about the Sun is proportional

to the cube of the planet’s mean distance from the Sun.

Let us prove this result for circular orbits. The gravitational force provides the necessary
centripetal force for circular motion:
m

θ
Replacing ω by 2π/T, we obtain M

A similar result is obtained for elliptical orbits, with the radius r replaced by the semi major axis a.

Problems
1. Two objects of masses 50kg and 500gm are situated at a distance 50cm away from each
other. Find (i) the magnitude of gravitational force acting between them, (ii) Acceleration
of lighter one due to the massive one (iii) Acceleration of massive one due to the lighter
one, (iv) which objects actually will move to the other one?

2. Suppose you are a sphere of radius 30cm and of mass 70kg. Calculate the acceleration
experienced by your friend standing 2m away from your center. (Assume that there is no
other force acting on both of you)

3. Determine the mass of the Earth from the period and radius of the moon’s orbit about the
earth. Period is 27.3 days and radius is 3.82 x 105km

4. A satellite orbits at a height of h=230km above the Earth’s surface. What is the period of
the satellite?

5. Calculate the minimum initial speed must a projectile have at the Earth’s surface to
escape from the earth?