114 學年度臺灣大學轉學生招生考試｜第 53 題
普通物理學(A)
- 0.57 hr
- 1.9 hr
- 5.7 hr
- 19 hr
- 57 hr
提示
Kepler's third law says T² is proportional to the cube of the orbital radius (measured from Earth's center). Scale the Moon's known period down to the satellite's much smaller orbit.
正確答案
正解：B
詳解
A. (A) 0.57 hr — incorrect. Too short; this is roughly a factor of 10 below the correct value and not consistent with the radius ratio.
B. (B) 1.9 hr — correct. The satellite's orbital radius is R+altitude = 6380+1500 = 7880 km = 7.88×10⁶ m. By Kepler's third law T_sat = T_Moon × (r_sat/r_Moon)^(3/2) = 655.2 hr × (7.88×10⁶/3.84×10⁸)^(3/2) ≈ 655.2 × (0.02052)^(1.5) ≈ 1.9 hr — the familiar ~90 min low-Earth-orbit period.
C. (C) 5.7 hr — incorrect. This overestimates the period; it does not follow from the 3/2-power scaling of the radius ratio.
D. (D) 19 hr — incorrect. Far too long for a low orbit just above the surface; a satellite this close orbits in roughly 90 minutes.
E. (E) 57 hr — incorrect. Grossly too long; this would correspond to an orbit much larger than the Moon's, the opposite of the situation.
