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 944.0 N on the earth weighs 918.0 N at the
north pole of Planet X and only 850.0 N at its equator. The
distance from the north pole to the equator is 1.882×10^{4}
km , measured along the surface of Planet X.

**Part A)** How long is the day on Planet X?

*t* = ? s

**Part B)** If a 4.400×10^{4} kg satellite
is placed in a circular orbit 2000 km above the surface of Planet
X, what will be its orbital period?

*T* = ? s

Answer #1

The mass of the atsronaut is:

m = 944/9.8 = 96.33 kg

The acceleration due to gravity of the planet will be:

g1 = 918/96.33 = 9.53 m/s^2

the acc. at the equator is

a = (918 - 850)/96.33 = 0.71 m/s^2

R = 2 x 1.882 x 10^4/pi = 1.198 x 10^7 m

A)We know that,

ac = w^2 R

w = sqrt (ac/R)

w = sqrt(0.71/1.198 x 10^7) = 2.43 x 10^-4 rad/s

2 pi f = 0.0077 => f = 3.88 x 10^-5

T = 1/f = 1/ 3.88 x 10^-5 = 25773.2 s

**Hence, T = 25773.2 s**

B)we need to know the mass of planet first.

M = g1xr^2/G

M = 9.53 x (1.198 x 10^7)^2/(6.67 x 10^-11) = 2.1 x 10^25 kg

we know from keplers law

T = 2 pi r sqrt (r/GM)

T = 2 x 3.14 x (11.98 x 10^6 + 2 x 10^6) sqrt ((11.98 x 10^6 + 2 x 10^6)/(6.67 x 10^-11)x(2.1 x 10^25))

T = 8771 s

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