Answer to Question #153919 in Astronomy | Astrophysics for NUR LIYANA IZZATI

Question

A satellite weighing 1600 kg revolves aroundt the Earth in an orbit radius three times that of the Earth.If the radius of the Earth is 6.4x10⁶m,what is the centripetal force acting on the satellite?

Answer 1

The centripetal force acting on the satellite is actually a Newton's gravity force:

$$F = G\dfrac{mM}{r^2}$$

where $m = 1600kg$ is the mass of the satellite, $M =6\times 10^{24}kg$ is the mass of the Earth, $r = 3R$ is the radius of the orbit, $R = 6.4\times10^6m$ is the radius of the Earth, and $G = 6.7\times10^{-11}m^3/(kg\cdot c^2)$ is the gravitational constant. Thus, obtain:

$$F = G\dfrac{mM}{9R^2}\\ F = 6.7\times10^{-11}\cdot \dfrac{1600\cdot 6\times 10^{24}}{9\cdot (6.4\times 10^6)^2} \approx 1745N$$

Answer. 1745 N.