Understanding Angular Momentum in Gravitational Systems

[Kitten207]

Homework Statement

The problem is stated here:

http://i53.tinypic.com/2wfl4jm.jpg


Please take a look.

 

[collinsmark]

Homework Statement

The problem is stated here:

http://i53.tinypic.com/2wfl4jm.jpg


Please take a look.

What do you think?

Just to get you started, The problem doesn't have much to do with angular momentum in particular. The title of the thread is "Angular Momentum Problem," but the problem more about gravitation, conservation of energy, and centripetal force.

Here are some starting hints:

• Part a): What is the gravitational potential difference between something on the surface of the asteroid (radius 4500 m) and something 400 meters above the surface?
• Part b): Use conservation of energy (for potential energy simply use mgh for situations here on Earth near the surface).
• Part c): What is the centripetal force on the spaceship (the force keeping it in orbit)? [There are two ways to formulate this force -- set them equal to each other. ]

 

[Kitten207]

I don't think I understand your first point, but this is what I got for it:

U_i = K_f
mgh = ½mv²
v= √2gh

What value do I use for g?

Thank you for the other two hints!

 

[Kitten207]

oh I'm a bit confused on the second part too actually. So:

U_i = K_f
mgh = ½mv²
v= √2gh
= √(2)(9.8m/s²)(400m)
=7840 m
I can't jump that fast...right? Is this correct?

 

[collinsmark]

I don't think I understand your first point, but this is what I got for it:

U_i = K_f
mgh = ½mv²
v= √2gh

What value do I use for g?

The gravitational P.E. = mgh is an approximation, and only applies if the height h is very, very small compared to the radius of the planet/asteroid. But that doesn't work so well here since you're jumping distance that is almost 1/10 of the entire radius of the planet. 1/10 is a significant portion of the radius, so I wouldn't use mgh for this part.

The more precise gravitational potential energy of spherical planet/asteroid can be found using the potential difference.

The gravitational potential of a spherical planet/asteroid can be found using

Potential = GM/r,​

where,
G = gravitational constant
M = mass of planet or asteriod
r = distance to the center of the planet/asteroid.
It is assumed that the potential is with respect a distance of infinity.

Calculate the gravitational potential at the asteroid's surface (one radius from its center). Then calculate the gravitational potential at a distance of Radius + 400 m from the asteroid's center. The difference between these two potentials is called the potential difference.

The difference in potential energy of an object of mass m between these two heights is the potential difference multiplied by m.

oh I'm a bit confused on the second part too actually. So:

U_i = K_f
mgh = ½mv²
v= √2gh
= √(2)(9.8m/s²)(400m)
=7840 m
I can't jump that fast...right? Is this correct?

You need to calculate v by solving part a) of the problem. By the time you get to part b) you should already know v.

Then once you get to part b), you need to solve for h, not v.

(But your choice of using mgh = ½mv² is good here . The height that you can jump here on Earth is minuscule compared to the radius of the Earth. So mgh = ½mv² is okay to use.)

 

[Kitten207]

thank you so much!