Question

NASA is expected to launch a 2, 600 kg satellite 450 km above the Earth’s surface. What is the expected velocity of the satellite when it orbits the Earth?

Answer 1

We can find the radius of the satellite as follows:

$$R=R_E+h=6.37\times10^6\ m+4.5\times10^5\ m=6.82\times10^6\ m.$$

We can find the speed of the satellite as follows:

$$F_c=F_g,$$ $$\dfrac{mv^2}{R}=\dfrac{GM_Em}{R^2},$$ $$v=\sqrt{\dfrac{GM_E}{R}}=\sqrt{\dfrac{6.67\times10^{-11}\ \dfrac{N\times m^2}{kg^2}\times5.98\times10^{24}\ kg}{6.82\times10^6\ m}},$$ $$v=7647\ \dfrac{m}{s}.$$