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Question: What will be the velocity of a satellite revolving around the earth at a height h above the surface ...

What will be the velocity of a satellite revolving around the earth at a height h above the surface of earth if radius of earth is R.
A. R2gR+h{{R}^{2}}\sqrt{\dfrac{g}{R+h}}
B. Rg(R+h)2R\dfrac{g}{{{(R+h)}^{2}}}
C. RgR+hR\sqrt{\dfrac{g}{R+h}}
D. RR+hgR\sqrt{\dfrac{R+h}{g}}

Explanation

Solution

To solve this question we will balance the gravitational force acting on the satellite with the centripetal force needed to keep the satellite in orbit. This way we will figure out what must be the velocity required for the satellite to stay in orbit and not fall to earth.

Formula used: Centripetal force: mv2r\dfrac{m{{v}^{2}}}{r}
Gravitational force: GMmr2\dfrac{GMm}{{{r}^{2}}}

Complete step by step answer:
First, we will calculate the gravitational force due to earth that will act on the satellite when it is at a distance h above the surface of earth.
Force of gravity is given as GMmr2\dfrac{GMm}{{{r}^{2}}}. Here G is the universal Gravitational constant. M and m are the masses of the two objects involved. Here M will be the mass of earth and m will be the mass of the satellite. ‘r’ is the distance between the centre of the planet and the satellite, here it will be R+h. So, we get the force of gravity as GMm(R+h)2\dfrac{GMm}{{{(R+h)}^{2}}}. We can write GMR2\dfrac{GM}{{{R}^{2}}} as g which is the acceleration due to gravity at the surface of earth. The force of gravity will then be written as mgR2(R+h)2\dfrac{mg{{R}^{2}}}{{{(R+h)}^{2}}}. When we balance it with the centripetal force, we get mgR2(R+h)2=mv2(R+h)⇒v2=gR2(R+h)\dfrac{mg{{R}^{2}}}{{{(R+h)}^{2}}}=\dfrac{m{{v}^{2}}}{(R+h)}\Rightarrow {{v}^{2}}=\dfrac{g{{R}^{2}}}{(R+h)}. And we get the velocity as v=Rg(R+h)v=R\sqrt{\dfrac{g}{(R+h)}}.

So, the correct answer is “Option C”.

Note: Take care that g is the acceleration due to gravity at the surface of the earth and we cannot take it to be the same at height h above the surface of earth. We could have take that assumption if it was given that h is very small when compared to the radius of earth since then the term R2(R+h)2\dfrac{{{R}^{2}}}{{{(R+h)}^{2}}} would be very close to one and the error would have been negligible.