a satellite orbiting earth has a tangential velocity of 5000 m/s. earths mass is 6×10²⁴ kg and its radius is…

Explanation:

Step1: Use the formula for the tangential velocity of a satellite

The formula for the tangential velocity (v) of a satellite in orbit around a planet of mass (M) at a distance (r) from the center of the planet is (v = \sqrt{\frac{GM}{r}}), where (G = 6.67\times10^{- 11}\space Nm^{2}/kg^{2}). We can re - arrange this formula to solve for (r): (r=\frac{GM}{v^{2}}).

Step2: Substitute the given values

We are given that (M = 6\times10^{24}\space kg), (v = 5000\space m/s), and (G = 6.67\times10^{-11}\space Nm^{2}/kg^{2}). Substitute these values into the formula: [ \begin{align*} r&=\frac{6.67\times 10^{-11}\times6\times 10^{24}}{(5000)^{2}}\ &=\frac{6.67\times6\times10^{-11 + 24}}{25\times10^{6}}\ &=\frac{40.02\times10^{13}}{25\times10^{6}}\ &=\frac{40.02}{25}\times10^{13-6}\ & = 1.6008\times10^{7}\space m\ \end{align*} ] This is the distance from the center of the Earth. To find the distance from the surface of the Earth, we need to subtract the radius of the Earth (R = 6.4\times10^{6}\space m) [h=r - R] [h=1.6008\times 10^{7}-6.4\times10^{6}] [h=(16.008\times10^{6}-6.4\times10^{6})\space m] [h = 9.608\times10^{6}\space m=9608000\space m]

Answer:

9,608,000 m