Question

NASA is expected to send a 2600-kg satellite 450 km above the earth’s surface.

(a) What is its radius?

(b) What speed will it have?

(c) What is its orbital period?

(d) What is its radial acceleration?

Answer 1

(a) We can find the radius of the satellite as follows:

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

(b) We can find the speed of the satellite as follows:

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

(c) We can find the orbital period of the satellite as follows:

$$v=\dfrac{2\pi R}{T},$$ $$T=\dfrac{2\pi R}{v}=\dfrac{2\pi\times6.82\times10^6\ m}{7647\ \dfrac{m}{s}}=5604\ s,$$ $$T=5604\ s\times\dfrac{1\ h}{3600\ s}=1.56\ h.$$

(d) We can find the radial (centripetal) acceleration as follows:

$$a_c=\dfrac{v^2}{R}=\dfrac{(7647\ \dfrac{m}{s})^2}{6.82\times10^6\ m}=8.57\ \dfrac{m}{s^2}.$$