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PHY 101 PHY 101 Past Questions Objective Question
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Question

A spacecraft is in circular orbit about the Earth at an altitude h of 400 km. Calculate the orbital speed and the centripetal acceleration if the spacecraft makes one revolution every 85 min.
Options
A 3.4 × 10³ m/s and 3.9 m/s²
Correct Answer
B 8.4 × 10³ m/s and 10.4 m/s²
C 4.4 × 10³ m/s and 9.8 m/s²
D 3 × 10³ m/s and 33 m/s²
Correct Answer

Option A is the correct answer.

Detailed Explanation

First convert period to seconds: T = 85 min * 60 s/min = 5100 s. Orbital radius r = Earth's radius (approx 6370 km) + 400 km = 6770 km = 6.77 × 10⁶ m. Orbital speed v = 2πr / T = 2π * 6.77e6 / 5100 ≈ (4.253e7)/5100 ≈ 8339 m/s ≈ 8.4 × 10³ m/s. That matches option B's speed. Centripetal acceleration a = v²/r = (8339)² / 6.77e6 ≈ 6.95e7/6.77e6 ≈ 10.26 m/s² ≈ 10.4 m/s². So option B is correct: 8.4 × 10³ m/s and 10.4 m/s².

Hint

Use v = 2πr/T and a = v²/r. Remember to convert units to SI.

Question Info

Course: PHY 101
Topic: PHY 101 Past Questions
Type: Objective
Difficulty: Not specified
Question 1177 of 1250
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