EK Physics

[IlyaR]

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I'm not understanding why B wouldn't be a correct answer. If MV^2/R = Constant gravitational force, wouldn't increasing M lower the V?

Thanks!

 

[monkeyvokes]

mv^2/r = force of gravity, as you said.
as the explanation says, set them equal:
mv^2/r = G mM/r^2 - m= satellite, M=planet
v^2 = G M/r

see, the mass of the satellite is gone?

 

[Romz]

^ Easy to see derivation.

I like to tackle these problems intuitively.

If I increase the the string between my center of rotation and a mass on the other end, it'll take more time for rotation.

I know in centripetal rotation, increasing the length of my string would mean DECREASING my ATTRACTIVE force Fg.

Already, I see A is the trick to this type of question. What is responsible for Attractive force? GMm/R^2. Aka Masses of Both Planets and radius between them. Orbital radius is a product of mv^2/r and I eliminate it. I want to focus on my attractive force.

From there, I see that a decrease in mass is the only way to decrease the attractive force between two objects and C is the answer.

Not gonna lie, I chose D because I didn't consider F=Gmm/R^2 and the inverse relationship between attractive force and radius between two bodies. My mistake, hopefully won't do it again.

It seems like a convoluted explanation, but I blanked on Gmm equation so I resorted to the centripetal force side of things. I find that type of theoretical knowledge helps in panicky situations. Stay calm and carry on.