Question

Mars, which has a radius of 3.4 × 106 m and a mass of 6.4 × 1023 kg, orbits the Sun, which has a mass of 2.0 × 1030 kg at a distance of 2.3 × 1011m.

Mar’s revolution around the sun is 687 days.

It takes 24 hours and 37 minutes to complete one rotation.









Which is greater, the tangential speed of Mars’s rotation or revolution?

Answer 1

Compute tangential speed = (2π × radius) / period.

Rotation:
- R = 3.4×10^6 m
- T_rot = 24 h 37 min = 24×3600 + 37×60 = 88,620 s
v_rot = 2π(3.4×10^6) / 88,620 ≈ 2.41×10^2 m/s ≈ 241 m/s

Revolution (orbit):
- orbital radius = 2.3×10^11 m
- T_rev = 687 days = 687×86,400 = 59,356,800 s
v_rev = 2π(2.3×10^11) / 5.93568×10^7 ≈ 2.44×10^4 m/s ≈ 24,400 m/s

Conclusion: Mars’ revolution (orbital) tangential speed (~2.4×10^4 m/s) is much greater than its rotational surface speed (~2.4×10^2 m/s) — about 100 times larger.