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) First, use the stone-throwing experiment to find the value of "g", the acceleration due to gravity. We know the stone's initial speed and its travel time (its mass is irrelevant), so use this kinematics equation: g = acceleration = (change in velocity) / time The velocity changes from +12 m/s (on the way up) to -12 m/s (on the way down), so: g = ((-12 m/s) - (+12 m/s)) / 5.00 s = -4.8 m/s² (The negative sign just means that gravity points down instead of up. Ignore it.) Next, use the equation that relates acceleration to force: acceleration = force / mass in our case: g = (gravity force) / m = (GMm/R²) / m = GM/R² where: G = Universal gravitational constant M = mass of Mongo R = radius of Mongo Solve for M: M = gR²/G They give you R; and we calculated "g" above, so plug in the numbers. 2) When you're orbiting in a circle, the graviational force is the same as the centripetal force. graviational force: GMm/r² (where r = 1.00×10^4 km) centripetal force: m?²r (where ? = ship's angular velocity in radians/sec) Set them equal (and divide the "m" out of both sides): GM/r² = ?²r Next, use the formula that relates the orbital period "T" to the angular velocity: ? = (2p)/T Plug that into the previous equation: GM/r² = ((2p)/T)²r Finally, just solve the above for "T"

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