Planet X rotates in the same manner as the earth, around an axis through its nor
ID: 1516220 • Letter: P
Question
Planet X rotates in the same manner as the earth, around an axis through its north and south poles, and is perfectly spherical. An astronaut who weighs 940.0 N on the earth weighs 918.0 N at the north pole of Planet X and only 857.0 N at its equator. The distance from the north pole to the equator is 1.886×104 km , measured along the surface of Planet X.
A) How long is the day on Planet X?
B) If a 4.600×104 kg satellite is placed in a circular orbit 1000 km above the surface of Planet X, what will be its orbital period?
Explanation / Answer
Let's find the radius:
arc length s = *r
1.886e4 km = /2 * r
r = 1.2006e4 km = 1.2006e7 m
mass of astronaut m = 940.0N / 9.8m/s² = 95.92 kg
so the gravitational acceleration on X is
g' = 918.0N / 95.92 = 9.571 m/s²
and the centripetal acceleration due to rotation at the equator is
a = (918.0 - 857.0)N / 95.92 kg = 0.64 m/s²
a) a = ²*r
0.64 m/s² = ² * 1.2006e7m
² = 5.3e-8 rad/s²
= 2.30e-4 rad/s
period T = 2 / = 27300.38 s = 7h 35m 38s
b) g' = GM / r²
9.571 m/s² = 6.674e11N·m²/kg² * M / (1.2006e7m)²
M = 2.06e25 kg
For "orbit", centripetal acceleration = gravitational acceleration, or
²r = (2/T)²r = 4²r / T² = v²/r = GM/r²
where the first four terms are all expressions for centripetal acceleration
and G = Newton's gravitational constant = 6.674e11 N·m²/kg²
and M = mass of central body
and r = orbit radius
and T = period
so
4²r / T² = GM/r²
T² = 4²r³ / GM
T = 2(r³ / GM)
T = 2( (1.2006e7 + 1e6)³m³ / (6.674e11 N·m²/kg² * 2.06e25kg) )
T = 7948.2 s = 2h 12m 28s