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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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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Answer 1 · 2016-10-26T19:22:11

$V_{B} = 7408 \frac{m}{s}$ $V_B=7408m/s$

Explanation:

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

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

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

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

For satellite A:

$V_{A} = \sqrt{g (\frac{R^{2}}{R + h_{A}})}$ $V_A = sqrt(g(R^2/(R + h_A)))$

$V_{A} = \sqrt{9.8 \frac{m}{s^{2}} \frac{{(6371000 m)}^{2}}{6371000 m + 539000 m}}$ $V_A = sqrt(9.8 m/s^2((6371000 m)^2)/(6371000 m + 539000 m))$

$V_{A} = 18131 \frac{m}{s}$ $V_A = 18131 m/s$

For satellite B:

$V_{B} = \sqrt{g (\frac{R^{2}}{R + h_{B}})}$ $V_B = sqrt(g(R^2/(R + h_B)))$

$V_{B} = \sqrt{9.8 \frac{m}{s^{2}} \frac{{(6371000 m)}^{2}}{6371000 m + 876000 m}}$ $V_B = sqrt(9.8 m/s^2((6371000 m)^2)/(6371000 m + 876000 m))$

$V_{B} = 7408 \frac{m}{s}$ $V_B = 7408 m/s$

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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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Explanation:

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