The definition of satellite is not necessarily restricted to the technological development of a satellite that orbits the Earth and transmits data. The more general definition is anything that is in orbit around a planet or moon. In fact, Earth is a satellite of the Sun and the Moon is a satellite of the Earth. You can use the equations that were provided, but keep in mind that you are dealing with uniform circular motion, so those equations apply as well.
Satellite Motion
\(\large\mathsf{ v_c = \sqrt{\frac{GM}{r}} }\)
\(\large\mathsf{ t_s = \frac{2\pi r^{\frac{3}{2}}}{\sqrt{GM}} }\)
OR
\(\large\mathsf{ t_s^2 = \frac{4\pi^2 r^3}{GM} }\)
Use these equations and what you know about uniform circular motion and satellite motion to solve these problems on your own. Then, come back and check your work. Be especially careful with your algebra!
| Problem | Picture | Given/Find | Equation | Solution |
|---|---|---|---|---|
| A spacecraft orbits Mars (mass= 6.40 x 1023 kg) in a circular orbit of radius 8.01 x 105 km. What is the period of the spacecraft? |
|
\(\small\mathsf{ M = 6.40 \times 10^{23} \text{ kg} }\) \(\small\mathsf{ r = 8.01 \times 10^{8} \text{ m}}\) \(\small\mathsf{ t_s = ? \text{ s}}\) |
\(\small\mathsf{ t_s = \frac{2\pi r^{\frac{3}{2}}}{\sqrt{GM}} }\) | \(\small\mathsf{ t_s = \frac{2\pi (8.01 \times 10^{8} \text{ m})^{\frac{3}{2}}}{\sqrt{(6.67 \times 10^{-11} \text{ m}^3 \text{/kg s}^2)(6.40 \times 10^{23} \text{ kg})}} }\) \(\small\mathsf{ t_s = \frac{2\pi \sqrt{(8.01 \times 10^{8} \text{ m})^3}}{\sqrt{(6.67 \times 10^{-11} \text{ m}^3 \text{/kg s}^2)(6.40 \times 10^{23} \text{ kg})}} }\) \(\small\mathsf{ t_s = 2.18 \times 10^{7} \text{ s} }\) |
| The moon of a planet is observed to have a nearly circular orbit (r = 4.00 x 105 km), and an orbital period of 21.5 days. What is the mass of the planet? |
|
\(\small\mathsf{ M = ? \text{ kg} }\) \(\small\mathsf{ r = 4.00 \times 10^{8} \text{ m}}\) \(\small\mathsf{ t_s = 21.5 \text{ days} \times \frac{86400 \text{ s}}{1 \text{ day}} = 1,857,600 \text{ s}}\) |
\(\small\mathsf{ t_s^2 = \frac{4\pi^2 r^3}{GM} }\) | \(\small\mathsf{ (1,857,600 \text{ s})^2 = \frac{4\pi^2 (4.00 \times 10^{8} \text{ m})^3}{(6.67 \times 10^{-11} \text{ m}^3 \text{/kg s}^2)(M)} }\) \(\small\mathsf{ M = \frac{4\pi^2 (4.00 \times 10^{8} \text{ m})^3}{(6.67 \times 10^{-11} \text{ m}^3 \text{/kg s}^2)(1,857,600 \text{ s})^2)} }\) \(\small\mathsf{ M = 1.10 \times 10^{25} \text{ kg} }\) |
| A satellite is placed in a circular orbit to observe the surface of Mars from an altitude of 9760 km. The equatorial radius of Mars is 3397 km. If the speed of the satellite is 1480 m/s, what is the magnitude of the centripetal acceleration of the satellite? |
|
\(\small\mathsf{ v_c = 1480 \text{ m/s} }\) \(\small\mathsf{ r = 1.3157 \times 10^7 \text{ m}}\) \(\mathsf{ a_c = ? \text{ m/s}^2 }\) |
\(\mathsf{ a_c = \frac{v^2}{r} }\) | \(\mathsf{ a_c = \frac{(1480 \text{ m/s})^2}{1.3157 \times 10^7 \text{ m}} }\) \(\mathsf{ a_c = 0.166 \text{ m/s}^2 }\) |
