Even though Kepler’s laws were able to explain the planetary motion, they failed to explain the forces responsible for it.

**Universal
Law of Gravitation**

Even
though Kepler’s laws were able to explain the planetary motion, they failed to
explain the forces responsible for it. It was Isaac Newton who analyzed Kepler’s
laws, Galileo’s observations and deduced the law of gravitation.

Newton’s
law of gravitation states that a particle of mass *M*_{1} attracts any other particle of mass *M*_{2} in the universe with an
attractive force. The strength of this force of attraction was found to be
directly proportional to the product of their masses and is inversely
proportional to the square of the distance between them. In mathematical form,
it can be written as:

Where is the unit vector from *M*_{1} towards *M*_{2}* *as shown in Figure 6.3, and G is the* *Gravitational constant that has the
value of 6.626 × 10^{−}^{11}
*N m*^{2} *kg*^{−}^{2} , and *r* is the distance between the two masses *M*_{1} and *M*_{2}
. In Figure 6.3, the vector denotes the gravitational force experienced by *M*_{2} due to *M*_{1}
. Here the negative sign indicates that the gravitational force is always
attractive in nature and the direction of the force is along the line joining
the two masses.

In cartesian
coordinates, the square of the distance is expressed as *y **r *^{2}* *=* *(*x*^{2}* *+ ^{2}*
z*^{2}). This is dealt in unit 2.

Consider
two point masses m_{1 }and m_{2} which are separated by a
distance of 10 meter as shown in the following figure. Calculate the force of
attraction between them and draw the directions of forces on each of them. Take
m_{1}= 1 kg and m_{2} = 2 kg

*Solution*

The
force of attraction is given by

From
the figure, r =10 m.

First,
we can calculate the magnitude of the force

It
is to be noted that this force is very small. This is the reason we do not feel
the gravitational force of attraction between each other. The small value of G
plays a very crucial role in deciding the strength of the force.

The
force of attraction ()
experienced by the mass m_{2} due to m_{1} is in the negative
‘y’ direction ie., rˆ =−*j*ˆ
. According to Newton’s third law, the mass m_{2} also exerts equal and
opposite force on m_{1}. So the force of attraction () experienced by m_{1}
due to m_{2} is in the direction of positive ‘y’ axis ie., *r*ˆ
= *j*ˆ .

The
direction of the force is shown in the figure,

Gravitational
force of attraction between m_{1} and m_{2}

=*
*−* *which confirms Newton’s third* *law.

As
the distance between two masses increases, the strength of the force tends to
decrease because of inverse dependence on *r*^{2}.
Physically it implies that the planet Uranus experiences less gravitational
force from the Sun than the Earth since Uranus is at larger distance** **from
the Sun compared to the Earth.

The
gravitational forces between two particles always constitute an action-reaction
pair. It implies that the gravitational force exerted by the Sun on the Earth
is always towards the Sun. The reaction-force is exerted by the Earth on the
Sun. The direction of this reaction force is towards Earth.

The
torque experienced by the Earth due to the gravitational force of the Sun is
given by

It implies that angular momentum is a constant vector. The angular momentum of the Earth about the Sun is constant throughout the motion. It is true for all the planets. In fact, this constancy of angular momentum leads to the Kepler’s second law.

The expression has
one inherent assumption that both *M*_{1} and *M*_{2}* *are treated as point masses. When it is* *said that Earth orbits around the Sun
due to Sun’s gravitational force, we assumed Earth and Sun to be point masses.
This assumption is a good approximation because the distance between the two
bodies is very much larger than their diameters. For some irregular and
extended objects separated by a small distance, we cannot directly use the
equation (6.3). Instead, we have to invoke separate mathematical treatment
which will be brought forth in higher classes.

However, this assumption about point masses holds even for small distance for one special case. To calculate force of attraction between a hollow sphere of mass M with uniform density and point mass m kept outside the hollow sphere, we can replace the hollow sphere of mass M as equivalent to a point mass M located at the center of the hollow sphere. The force of attraction between the hollow sphere of mass M and point mass m can be calculated by treating the hollow sphere also as another point mass.

Essentially the entire mass of the hollow
sphere appears to be concentrated at the center of the hollow sphere. It is
shown in the Figure 6.5(a).

There
is also another interesting result. Consider a hollow sphere of mass M. If we
place another object of mass ‘m’ inside this hollow sphere as in Figure 6.5(b),
the force experienced by this mass ‘m’ will be zero. This calculation will be
dealt with in higher classes.

The
triumph of the law of gravitation is that it concludes that the mango that is
falling down and the Moon orbiting the Earth are due to the same gravitational
force.

Newton
considered the orbits of the planets as circular. For circular orbit of radius
r, the centripetal acceleration towards the center is

Here
v is the velocity and r, the distance of the planet from the center of the
orbit (Figure 6.6).

The
velocity in terms of known quantities r and T, is

Here
T is the time period of revolution of the planet. Substituting this value of v
in equation (6.4) we get,

Substituting
the value of ‘a’ from (6.6) in Newton’s second law, *F* = *ma*,
where ‘m’ is the mass of the planet.

From
Kepler’s third law,

By
substituting equation 6.9 in the force expression, we can arrive at the law of
gravitation.

Here
negative sign implies that the force is attractive and it acts towards the
center. In equation (6.10), mass of the planet ‘m’ comes explicitly. But Newton
strongly felt that according to his third law, if Earth is attracted by the
Sun, then the Sun must also be attracted by the Earth with the same magnitude
of force. So he felt that the Sun’s mass (M) should also occur explicitly in
the expression for force (6.10). From this insight, he equated the constant 4π^{2}*k to GM *which turned out to be the law of
gravitation

Again
the negative sign in the above equation implies that the gravitational force is
attractive.

In the above discussion we assumed that the orbit of the planet to be circular which is not true as the orbit of the planet around the Sun is elliptical. But this circular orbit assumption is justifiable because planet’s orbit is very close to being circular and there is only a very small deviation from the circular shape.

**EXAMPLE 6.2**

Moon
and an apple are accelerated by the same gravitational force due to Earth.
Compare the acceleration of the two.

The
gravitational force experienced by the apple due to Earth

Here
M_{A }– Mass of the apple, M_{E}– Mass of the Earth and R –
Radius of the Earth.

Equating
the above equation with Newton’s second law,

Simplifying
the above equation we get,

Here
*a _{A}* is the acceleration of
apple that is equal to ‘g’.

Similarly
the force experienced by Moon due to Earth is given by

Here
R_{m}- distance of the Moon from the Earth, M_{m} – Mass of the
Moon

The
acceleration experienced by the Moon is given by

The
ratio between the apple’s acceleration to Moon’s acceleration is given by

From
the Hipparchrus measurement, the distance to the Moon is 60 times that of Earth
radius. R_{m} = 60R.

The
apple’s acceleration is 3600 times the acceleration of the Moon.

The
same result was obtained by Newton using his gravitational formula. The apple’s
acceleration is measured easily and it is 9.8 m s^{−2} . Moon orbits
the Earth once in 27.3 days and by using the centripetal acceleration formula,
(Refer unit 3).

which
is exactly what he got through his law of gravitation.

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11th Physics : Gravitation : Universal Law of Gravitation |

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