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=6:67×10−11Nm2=kg2
, Mass of the earth is
5:98×1024
kg

Answer 1

Answer on Question#40854 – Physics – Mechanics

A 2000 kg satellite orbits the earth at a height of 300 km. What is the speed of the satellite and its period? Take $G = 6:67 \times 10^{-11} \, \text{Nm}^2 \cdot \text{kg}^2$, Mass of the earth is $5:98 \times 10^4 \, \text{kg}$

Solution:

$$M = 5.98 \times 10^{24} \, \text{kg} - \text{mass of Earth}$$ $$G = 6.67 \times 10^{-11} \, \text{N} \cdot \frac{\text{m}^2}{\text{kg}^2} - \text{gravitational constant};$$ $$R = 6.3 \times 10^6 \, \text{m} - \text{radius of Earth};$$ $$h = 300 \, \text{km} - \text{satellite orbits heigth above the Earth};$$

Consider a satellite with mass $m$ orbiting a central body with a mass of mass $M_{\text{earth}}$. If the satellite moves in circular motion, then the net centripetal force acting upon this orbiting satellite is given by the relationship

$$F_{\text{net}} = m \frac{v^2}{R}$$

This net centripetal force is the result of the gravitational force that attracts the satellite towards the central body and can be represented as

$$F_{\text{grav}} = G \frac{mM}{R^2}$$

Since $F_{\text{net}} = F_{\text{grav}}$, the above expressions for centripetal force and gravitational force can be set equal to each other. Thus,

$$\begin{array}{c} (1) = (2): \\ m \frac{v^2}{R} = G \frac{mM}{R^2} \end{array}$$

Observe that the mass of the satellite is present on both sides of the equation; thus it can be canceled by dividing through by $m$. Then both sides of the equation can be multiplied by $R$, leaving the following equation.

$$v^2 = \frac{GM}{R}$$ $$v = \sqrt{\frac{GM}{R + h}} = \sqrt{\frac{6.67 \times 10^{-11} \, \text{N} \cdot \frac{\text{m}^2}{\text{kg}^2} \cdot 5.98 \times 10^{24} \, \text{kg}}{6.3 \times 10^6 \, \text{m} + 300 \times 10^3 \, \text{m}}} = 7774 \, \frac{\text{m}}{\text{s}}$$

The equation that is useful in describing the motion of satellites is Newton's form of Kepler's third law. The period of a satellite (T) and the mean distance from the central body $(R + h)$ are related by the following equation:

$$T = 2\pi \sqrt{\frac{(R + h)^3}{GM}} = 2 \cdot 3.14 \sqrt{\frac{\frac{T^2}{R^3} = \frac{4\pi^2}{GM}}{6.67 \times 10^{-11} \, \text{N} \cdot \frac{\text{m}^2}{\text{kg}^2} \cdot 5.98 \times 10^{24} \, \text{kg}}} = 5332 \, \text{s}$$ $$= 1.48 \, \text{hours}$$

Answer: speed of the satellite: $v = 7774 \, \frac{\text{m}}{\text{s}}$;

period of the satellite: $T = 1.48$ hours.