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.

For circular motion, the centripetal force is F = mv²/r, so the orbital radius is r = mv²/F. Substituting the values gives r = (1,160)(7,446²)/8,955 ≈ 7.182 × 10⁶ m. Subtract Earth’s radius: 7.182 × 10⁶ − 6.38 × 10⁶ ≈ 8.02 × 10⁵ m, so the height is choice B, 8.02 × 10⁵ m.

AI-written and checked against the verified answer.