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,

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

\(M=2.03\times10^{18}\; kg\)