Question

A 2000 kg satellite orbits the earth at a height of 300 km. What is the speed of the satellite and its period? Take G = 6:67 x 10-11Nm2=kg2
, Mass of the earth is 5:98 x 1024kg

Answer 1

Answer on Question #51081, Physics, Mechanics | Kinematics | Dynamics

Question:

A 2000 kg satellite orbits the earth at a height of 300 km. What is the speed of the satellite and its period? Take $G = 6:67 \times 10^{-11} \, \text{Nm}^2 \, \text{kg}^2$, Mass of the earth is 5:98 x 1024 kg.

Answer:

Newton's second law of motion:

$$\frac{m v^2}{R + h} = \frac{G M m}{(R + h)^2}$$

where $\frac{v^2}{R + h}$ is centripetal acceleration, $R$ is radius of Earth

Therefore, speed of motion equals:

$$v = \sqrt{\frac{G M}{R + h}} = 7730 \, \frac{m}{s} = 7.73 \, \frac{\text{km}}{\text{s}}$$

Period equals:

$$T = \frac{2 \pi (R + h)}{v} = 2 \pi \sqrt{\frac{(R + h)^3}{G M}} \cong 5420 \, \text{s}$$

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