Angular Acceleration and Moment Arm

[RoboNerd]

Homework Statement

[A rocket has landed on Planet X, which has half the radius of Earth. An astronaut onboard the rocket weighs twice as much on Planet X as on Earth. If the escape velocity for the rocket taking off from Earth is v , then its escape velocity on Planet X is

a) 2 v
b) (√2)v
c) v
d) v/2
e) v/4

C is the answer

Homework Equations

so I know the following

0.5 * m * (vEscape)^2 - [ (G * m * mEarth)/r ] = 0.

The Attempt at a Solution

[/B]
I know that my g on this planet is going to be twice as much so the radius of this planet is decreased by a factor of (1 / sqrt(2) ) [via Newton's law of gravitation]. Thus, I think that planet X's radius should be changed to (1 / sqrt(2) ) * radius of Earth, and that would create a new escape velocity different from v.

Could anyone please explain how C is the right answer?

Thanks in advance.

 

[Orodruin]

Escape velocity is related to the gravitational potential, not to the gravitational force.

 

[RoboNerd]

Could you please explain how you would have solved this? Thanks!

 

[haruspex]

Could you please explain how you would have solved this? Thanks!

What is the gravitational potential at the surface of a uniform sphere, mass m, radius r?

 

[Orodruin]

I know that my g on this planet is going to be twice as much so the radius of this planet is decreased by a factor of (1 / sqrt(2) ) [via Newton's law of gravitation]. Thus, I think that planet X's radius should be changed to (1 / sqrt(2) ) * radius of Earth, and that would create a new escape velocity different from v.

By the way, this makes absolutely no sense. Why would you change the parameters of the problem? Nowhere is it stated that the planet has the same mass as Earth.

 

[RoboNerd]

well i think they do say something about the mass of planet s when they say that the g is different from the one on Earth by a factor of two.

gravitational potential at Earth's surface is (-G * mE * mObject)/r

 

[Orodruin]

well i think they do say something about the mass of planet s when they say that the g is different from the one on Earth by a factor of two.

They are, but your statement about the radius would only have been valid if the planet had the same mass as the Earth. So what is the gravitational potential at the surface of the planet? How does it relate to the surface acceleration?

 

[RoboNerd]

(G * massEarth * massObjectAtSurface)/ radiusEarth = gravitational potential at surface.

I do not know how it relates to surface acceleration.

 

[Orodruin]

(G * massEarth * massObjectAtSurface)/ radiusEarth = gravitational potential at surface.

I do not know how it relates to surface acceleration.

What is the expression for the surface acceleration?

 

[RoboNerd]

( G * mE ) / r^2 = surface acceleration (g)

 

[Orodruin]

And therefore ...

 

[RoboNerd]

We do not have enough information to make a valid conclusion. Thanks!

 

[haruspex]

We do not have enough information to make a valid conclusion. Thanks!

Wrong.

 

[Orodruin]

We do not have enough information to make a valid conclusion. Thanks!

You do have enough information, you are just not using it right.

 

[RoboNerd]

OK... what would you suggest I do to get this right?

What would be the next step?

 

[haruspex]

OK... what would you suggest I do to get this right?

What would be the next step?

Create a variable for the mass of the rocket. Using your equation in post #10, write an expression for the weight of the rocket on Earth. Create another variable for the mass of planet X. Knowing the radius of planet X, write an expression for the weight of the rocket on planet X.
What equation can you now write?

 

[RoboNerd]

weight of the rocket on Earth = mg = m * [ (G * mE) / (rE^2) ]

weight of rocket on planet x = mg = m [ (G * mX) / (1/4 * rE)^2 ].

I multiply the first equation by two and set it equal to the second one

2* m * [ (G * mE) / (rE^2) ] = m [ (G * mX) / (1/4 * rE)^2 ]

I solved for mX and I get 1/4 * mE.

right?

 

[Nathanael]

I solved for mX and I get 1/4 * mE.

right?

That is the right idea, but be careful with your math:

2* m * [ (G * mE) / (rE^2) ] = m [ (G * mX) / (1/4 * rE)^2 ]

If you cancel all the similar constants out of this equation you get 2m_E = m_x/(1/4) = 4m_x so then m_x = 1/2 m_E

The next question is, how does the escape velocity of a planet depends on the mass/radius of the planet?
(Orodruin's post #2 should help).

 

[RoboNerd]

Thanks so much for the help! I got it! I wrote an expression for escape velocity on planet x, substituted terms, and the new coefficients canceled!

I really appreciate this!