Question

A large telescope of mass 8410 kg is in circular orbit around the earth, making one revolution every 927 minutes. What is the magnitude of the gravitational force exerted on the satellite by the earth? G = 6.67*10^-11 N. m^2 kg^-2, M_earth = 6.0*10^24 KG

Answer 1

Using the Kepler's third law and Newton's law of universal gravitation

$$T^2=\frac{4\pi^2 r^3}{GM}, \quad F=G\frac{mM}{r^2}$$

we obtain

$$F=GmM\left(\frac{4\pi^2}{GMT^2}\right)^{2/3}$$

$$F=6.67*10^{-11}*8410*6.0*10^{24}\\\left(\frac{4\pi^2}{6.67*10^{-11}*6.0*10^{24}*(927*60)^2}\right)^{2/3}\\ =3382\:\rm N$$