Satellite Orbital Analysis

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Unit: Uniform Circular Motion and Gravitation

Prerequisites

Gravitational Field Orbital Motion and Kepler's Third Law

Later Topics

None

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Part 1

A satellite orbits Earth at an altitude of $h_1 = 422$ km above the surface. Given Earth's mass $M = 5.97\times 10^{24}\text{ kg}$ , Earth's radius $R = 6370000$ m, and the gravitational constant $G = 6.67\times 10^{-11}\text{ N}\text{m}²/\text{kg}²$ , calculate the orbital speed $v_1$ and the orbital period $T_1$ of the satellite.

Correct!

Solution:

Solution

The orbital radius from Earth's center is: $r_1 = R + h_1 = 6370000 + 422000 = 6790000 \text{ m}$

For a circular orbit, the gravitational force provides the centripetal force. Setting these equal: $\frac{GMm}{r_1^2} = \frac{mv_1^2}{r_1}$

Solving for the orbital speed $v_1$ : $v_1 = \sqrt{\frac{GM}{r_1}}$

Substituting values: $v_1 = \sqrt{\frac{(6.67e-11) \cdot (5.97e24)}{6790000}} \approx 7657 \text{ m/s}$

The orbital period is the circumference divided by the speed: $T_1 = \frac{2\pi r_1}{v_1}$

Substituting values: $T_1 = \frac{2 \cdot \pi \cdot 6790000}{7657} = 5573.5 \text{ s} = 92.9 \text{ min}$

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