Newtons Gravitational Force

Around 50 years after Kepler's laws had been introduced, Isaac Newton proposed that every object in the universe attract every other object with a force which is,

• Inversely proportional to the square of their separation and,
• Directly proportional to the product of their masses.

Thus,

$$F={G m_1 m_2\over r^2}$$

where,

F         - The gravitational force between the objects.

r         - The separation between the two objects

m1, m2 - The masses of the two objects

G        - The proportionality constant known as the universal gravitational constant. The value equals to $$6.67\times 10^{-11}N m^2kg^{-2}$$.

This equation can be used to prove Kepler's 3rd law.

The centripetal force; for an object of mass m, moving in a circular orbit around the earth with a speed v, is given by the equation;

$$F_c = {mv^2\over r}$$

This is also the gravitational force acting on the object. If M is the mass of the earth,

$$F ={ G Mm\over r^2}$$

Thus,

$${mv^2\over r} ={G Mm\over r^2}$$

$${v^2} = {GM\over r}$$

​As the object is moving in a circular orbit, the total distance traveled during one revolution is $$2\pi r$$.

​If the time taken for a complete revolution is T, then

$$T ={ 2\pi r\over v}$$

$$T^2 = {4\pi^2 r^2\over v^2}$$

By substituting the equation for speed,

$$T^2={4\pi^2 r^2\over {GM\over r}}$$

$$T^2={4\pi^2\over GM} r^3$$

This shows that, $$T^2 \propto r^3$$​  (Keplers' Third Law)

​Example 1:

A satellite of mass 66 kg is in an orbit around the earth at a distance 3200 km above its surface. If the mass of the earth is $$6.0\times10^{24}\;kg$$ and its radius is 6400 km, find the gravitational force between the earth and the satellite. $$G=6.67\times 10^{-11}N m^2kg^{-2}$$

​The mass of the earth (M) = $$6.0\times10^{24}\;kg$$

The mass of the satellite (m) = 66 kg

The distance from the center of the earth to the staellite(r) = 3200 km+6400 km=9600 km

Using the equation,

$$F={GMm\over r^2}$$

$$F=({6.67\times 10^{-11}\;N m^2kg^{-2})(6.0\times10^{24}\;kg)(66\; kg)\over(9600\times10^3\;m)^2}$$

$$F=286.60\;N$$

Example 2:​

​A space craft orbitting the earth is $$1\times10^7m$$  ​away from the center of the earth. It takes 90 mins to complete an orbit. What is the mass of the earth? $$G=6.67\times 10^{-11}N m^2kg^{-2}$$

The distance to the space craft from the center of the earth (r) $$=1\times10^7m$$

The time taken for a complete orbit (T) $$=90\;min=90\times60\;s=5400\;s$$

Using the equation, $$T^2={4\pi^2\over GM} r^3$$

By substituting the values

$${(5400\;s)^2}={4\pi^2\over6.67\times10^{-11}\;Nm^2kg^{-2}\times M}\times(1\times10^7m)^3$$

By isolating M,

$${M}={4\pi^2\over6.67\times10^{-11}\;Nm^2kg^{-2}\times (5400\;s)^2}\times(1\times10^7m)^3$$

$$M=2.03\times10^{18}\; kg$$