Question

Calculate the rotational kinetic energy (in J) of Venus on its axis. (Assume its mass to be 4.90 ✕ 1024 kg, the period about its axis to be 5830 h, and its diameter to be 12100 km.)
(b)
What is the rotational kinetic energy (in J) of Venus in its orbit around the Sun? (Assume its distance from the sun to be 1.08 ✕ 108 km and its period about the sun to be 225 days.)

Answer 1

We treat Venus as a uniform sphere for its spin (I = (2/5)MR^2) and as a point mass for its orbital motion (KE = 1/2 Mv^2, with v = 2πr/T).

Given:
- M = 4.90×10^24 kg
- spin period Tspin = 5830 h = 5,830×3600 = 2.0988×10^7 s
- diameter = 12,100 km → R = 6.05×10^6 m
- orbital radius r = 1.08×10^8 km = 1.08×10^11 m
- orbital period Torb = 225 d = 225×86400 = 1.944×10^7 s

(a) Spin (rotational about axis)
I = (2/5) M R^2 = 0.4(4.90×10^24)(6.05×10^6)^2 ≈ 7.17×10^37 kg·m^2
ω = 2π/Tspin = 2π / 2.0988×10^7 s ≈ 2.995×10^−7 s^−1
Kspin = 1/2 I ω^2 ≈ 0.5·7.17×10^37·(2.995×10^−7)^2 ≈ 3.22×10^24 J

(b) Orbital (about the Sun)
v = 2πr / Torb = 2π·1.08×10^11 / 1.944×10^7 ≈ 3.49×10^4 m/s
Korb = 1/2 M v^2 ≈ 0.5·4.90×10^24·(3.49×10^4)^2 ≈ 2.98×10^33 J

So:
- Rotational KE about its axis ≈ 3.2×10^24 J
- Rotational (orbital) KE about the Sun ≈ 3.0×10^33 J

(The orbital kinetic energy is about 10^9 times larger than the spin kinetic energy.)