# Homework Help: Angular Acceleration and Moment Arm

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1. Mar 19, 2016

### RoboNerd

1. The problem statement, all variables and given/known data
[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

2. Relevant equations
so I know the following

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

3. The attempt at a solution

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.

2. Mar 19, 2016

### Orodruin

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

3. Mar 19, 2016

### RoboNerd

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

4. Mar 19, 2016

### haruspex

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

5. Mar 19, 2016

### Orodruin

Staff Emeritus
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.

6. Mar 20, 2016

### 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 earths surface is (-G * mE * mObject)/r

7. Mar 20, 2016

### Orodruin

Staff Emeritus
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?

8. Mar 20, 2016

### RoboNerd

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

I do not know how it relates to surface acceleration.

9. Mar 20, 2016

### Orodruin

Staff Emeritus
What is the expression for the surface acceleration?

10. Mar 20, 2016

### RoboNerd

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

11. Mar 21, 2016

### Orodruin

Staff Emeritus
And therefore ...

12. Mar 22, 2016

### RoboNerd

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

13. Mar 22, 2016

### haruspex

Wrong.

14. Mar 23, 2016

### Orodruin

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

15. Mar 23, 2016

### RoboNerd

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

What would be the next step?

16. Mar 23, 2016

### haruspex

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?

17. Mar 25, 2016

### 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?

18. Mar 25, 2016

### Nathanael

That is the right idea, but be careful with your math:
If you cancel all the similar constants out of this equation you get 2mE = mx/(1/4) = 4mx so then mx = 1/2 mE

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).

19. Mar 25, 2016

### 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!