Question

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly \[36,000\text{ }km\] from the surface of the earth. What is the potential due to earth's gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = \[6.0\times {{10}^{24}}kg\], radius = \[6400\text{ }km\].

Answer 1

Hint: Applying the formula of the gravitational potential energy where the work done against the gravity or work done to keep a body within its gravitational sphere i.e. bringing an object from infinity, the gravitational potential energy formula used is:
\[V=\dfrac{-\text{ }GMm}{H}\]
where M is the mass of the earth, m is the mass of the satellite, G is the gravitational force, H is the radius or the length from earth's core to the object.

Complete step by step answer:
Now let us see the diagram is drawn below, we can see that the total distance from the earth's core to the satellite is both the sum of the earth's care radius and the distance from earth's surface to that of the satellite meaning the total radius or length is \[R+r\].
\[R+r=36000\text{ }km+6400\text{ }km\]
\[\Rightarrow R+r=42600\text{ }km\]
Converting the distance from km to m, we get the value of the total distance H as:
\[42600\text{ }km\times 1000=4.24\times {{10}^{7}}m\].
Now as the mass of the satellite is not given hence, we take the mass of the satellite as 1 and after taking the mass of the satellite as 1, we have the values of the mass of the earth as:
\[M=6.0\times {{10}^{24}}kg\]
And the value of the gravitational constant is taken as \[G=6.67\times {{10}^{^{-}11}}\].
After this we place all the values in the formula with mass of the satellite as 1, we get the value of the potential energy on the site of the satellite as:
\[V=\dfrac{-\text{ }GMm}{H}\]
\[\Rightarrow V=\dfrac{-\text{ }6.67\times {{10}^{^{-}11}}\times 6.0\times {{10}^{24}}\times 1}{4.24\times {{10}^{7}}m}\]
Simplifying the values of the given terms carefully we get the value of the potential energy as:
\[V=-9.4\times {{10}^{6}}J/kg\]
Therefore, the potential energy due to the earth on the site of the satellite is given as \[-9.4\times {{10}^{6}}J/kg\].

Note:
The negative symbol in the potential energy indicates the bound state of the earth's gravitational force on the satellite meaning the amount of energy required to keep the satellite within the line of escape. The line or circumference of escape is the final line until which objects are within the gravitational force beyond that they escape the body's gravitational force and start moving away from the body into space.