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 11.0 m/s returns to the ground in 9.00 s ; the circumference of Mongo at the equator is 4.00×105 km ; and there is no appreciable atmosphere on Mongo.

Part A

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

Part B

If the Aimless Wanderer goes into a circular orbit 3.00×104 km above the surface of Mongo, how many hours will it take the ship to complete one orbit?

Explanation / Answer

Time to for stone to reach max height is
t =u/g
time to go up and down is twice this
t =2 x u/g
g = 2 x u/t
g = 2 x 11/9
g = 22/9
g = 2.44 ms-2

gp= GMp/Rp^2
Circumference of planet = 2 x pi x Rp
Rp = circum/2 x 3.142
Rp = 4 x 10^5 km/6.284
Rp = 10^8/6.284
Rp = 6.36 x 10^7 m
The radius of the planet is 6.36x 10^7 m

Rp^2 = (6.36 x 10^7)^2 = 40.56 x 10^14 m^2

gp= GMp/Rp^2
Solving for Mp gives:-
Mp=(gp x Rp^2)/G
Mp =(2.44 x 40.56 x 10^14)/6.62 x 10^-11
Mp=14.95 x10^25 kg

Mp = 14.95 x 10^25 kg

B)
Use: T = 2 x pi x Squ root of [r^3/GMp]
Where r = 3 x 10^7 m Plus the radius of the planet
r = 3 x 10^7 +6.36x 10^7
r = 9.36 x 10^7
r^3 = 82 x 10^22
T = 2 x pi x Squ root of [r^3/GMp]
T = 6.284 x Squ root of [82 x 10^22/(6.62 x 10^-11 x 14.95x 10^25]
T = 6.284 x Squ root of [0.082 x 10^8]
T = 6.284 x 2869.3
T = 18019 seconds
T = 5.005 Hours

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