Planet X rotates in the same manner as the earth, around an axis through its nor
ID: 1470491 • 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 950.0 N on the earth weighs 916.0 N at the north pole of Planet X and only 840.0 N at its equator. The distance from the north pole to the equator is 1.888×104 km , measured along the surface of Planet X.
Part A
How long is the day on Planet X?
Part B
If a 4.200×104 kg satellite is placed in a circular orbit 3000 km above the surface of Planet X, what will be its orbital period?
Explanation / Answer
A) The centrifugal force is the difference between the north pole mass and the equator mass.
Fc = m(omega^2)r. we do not have r, but we have a quarter circumference, which we can use to solve for it (assuming the planet is a circle, which it won't quite be, but whatever). we can use astro's earth weight to come up with his mass.
The day is just 2 pi radians / omega.
B)
"The distance from the north pole to the equator is 1.885×104km"
s = 1.885*104 km = *r = /2 * r
r = 1.20*104 km
gravitational acceleration on X is
a = 9.8m/s² * 916/950 = 9.45 m/s²
and a = GM / r², so
M = ar² / G = 9.45m/s² * (1.2*107m)² / 6.674*10-11N·m²/kg² = 2.038*1025 kg
For "orbit", centripetal acceleration = gravitational acceleration, or
²r = (2/T)²r = 4²r / T² = GM/r² T² = 4²r³ / GM
where G = Newton's gravitational constant = 6.674*10-11 N·m²/kg²
and M = mass of central body
and r = orbit radius
Plugging in values, find
T² = 5*106 s²
T = 2236.06 s