Q.A satellite revolves around the Earth in a circular orbit at a height of $300\ \text{km}$ above the surface. Taking $GM = 4.00 \times 10^{14}\ \text{m}^3/\text{s}^2$ and $R = 6400\ \text{km}$, find

Equate the gravitational pull of the Earth on the satellite to the centripetal force needed to keep it in its circular orbit.

Consider a satellite of mass $m$ revolving in a circular orbit of radius $(R+h)$ around the Earth (Earth's radius $R$, mass $M$, height of orbit above surface $h$).

For the satellite to remain in a stable circular orbit, the gravitational force of attraction provided by the Earth must exactly supply the centripetal force required for circular motion:

$\frac{GMm}{(R+h)^2}=\frac{mv_c^2}{R+h}$

Cancelling $m$ and one factor of $(R+h)$ from both sides:

$v_c^2=\frac{GM}{R+h}$

$v_c=\sqrt{\frac{GM}{R+h}}$

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