Question

A 1,160 kg satellite orbits Earth with a tangential speed of 7,446 m/s. If the satellite experiences a centripetal force of 8,955 N, what is the height of the satellite above the surface of Earth? Recall that Earth’s radius is 6.38 × 106 m and Earth’s mass is 5.97 × 1024 kg.

Answer 1

8.02*10⁵ m

Explanation:

The height of the satellite above the surface of Earth is 8.02\times 10^{5}m

Explanation:

Given

Mass of the satellite, m= 1160 kg

tangential speed , v = 7446 m/s

Centripetal force , F = 8955 N

Radius of earth , R= 6.38\times 10⁶ m

Let height of satellite above surface of the Earth be H

Centripetal force on satellite is given by

F= mv²/ R+H

=> H= mv²/ F - R

=> H= ( 1160*7446⁶/8955 - 6.38 * 10⁶ ) m

Thus, the height of the satellite above the surface of Earth is 8.02*10⁵ m.

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