# Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 539 km above the earth's surface, while that for satellite B is at a height of 876 km. How do you find the orbital speed for satellite A and satellite B?

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Oct 27, 2016

#### Answer:

${V}_{B} = 7408 \frac{m}{s}$

#### Explanation:

To do this problem, you need the Earth's radius $R = 6371 k m$

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration ${V}^{2} / \left(R + h\right)$ must equal the acceleration due to gravity at that distance from the center of the earth $g \left({R}^{2} / {\left(R + h\right)}^{2}\right)$

${V}^{2} / \left(R + h\right) = g \left({R}^{2} / {\left(R + h\right)}^{2}\right)$

$V = \sqrt{g \left({R}^{2} / \left(R + h\right)\right)}$

For satellite A:

${V}_{A} = \sqrt{g \left({R}^{2} / \left(R + {h}_{A}\right)\right)}$

${V}_{A} = \sqrt{9.8 \frac{m}{s} ^ 2 \frac{{\left(6371000 m\right)}^{2}}{6371000 m + 539000 m}}$

${V}_{A} = 18131 \frac{m}{s}$

For satellite B:

${V}_{B} = \sqrt{g \left({R}^{2} / \left(R + {h}_{B}\right)\right)}$

${V}_{B} = \sqrt{9.8 \frac{m}{s} ^ 2 \frac{{\left(6371000 m\right)}^{2}}{6371000 m + 876000 m}}$

${V}_{B} = 7408 \frac{m}{s}$

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