Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As ch

Question

Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: a 2.50-kg stone thrown upward from the ground at 15.0 m/s returns to the ground in 9.00 s ; the circumference of Mongo at the equator is 2.00×10^5 KM; and there is no appreciable atmosphere on Mongo.

1-The starship commander, Captain Confusion, asks for the following information: what is the mass of Mongo?

2-If the Aimless Wanderer goes into a circular orbit 2.00×10^4 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

Explanation / Answer

1. We can use kinematics to find the acceleration of gravity at the surface of Mongo:

vo = 15.0m/s
v = -15.0m/s (stone returns to ground)
t = 9.00s
a = ?

v = vo + at
-15.0m/s = 15.0m/s + a(9.0s)
a = (-30m/s)/9.0s = 3.333 m/s^2 (magnitude only)

Now let's use this value along with newton's 2nd to find the mass of the planet.

F = m*a

Gm*M/r^2 = m*a

m's cancel

GM/r^2 = a

convert km to m and we get

M = a*r^2/G = (3.33m/s^s)(2.0*10^8m)/(6.67*10^-11) = 1.999*10^27 kg

2. If we are in orbit 2.0*10^4km above the surface, then we are (2*10^5km + 2*10^4km)= 2.2*10^5km from the center. So:

GmM/r^2 = mv^2/r

m's and one r cancel:

GM/r = v^2

let's solve for v (again convert km to m):

v^2 = (6.67*10^-11)(1.999*10^27)/(2.2*10^8m)

v = 24,618m/s

If v = (one circumference)/Period = 2r/T

Then

T = 2r/v = 2(2.2*10^8m)/(24,618m/s) = 56,149s

or

15.6 hours

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