Energy of a particel in a gravitational field

[Gogsey]

4. A particle is launched vertically from the surface of the earth, and rises to a height of 2RE above the surface
before falling back. Ignore air resistance entirely.
a) Find the initial speed of the particle. (Look up the necessary numbers.)
b) This is a problem of motion in one dimension, with a force that depends on position only. Find the time
required to reach the turning point (maximum height).
(RE is just the radius of the earth).

This seeems so simple that I must be approaching this all wrong but here goes.

a). Putting the origin at the Earth's surface, we get that mvo^2/2 = mgh, where h = 2RE.
Therefore rearranginf for vo we get vo = 2sqrt(gRE).
I'm skeptical whether this is right or not because this is a level 2 mechanics course, and it just seems to easy for the course and the prof.

b), We known that U(r) - U(ro) = integral from ro to r of F dr, and thet U(ro) is zero, and
U(r) = 2mgRE.
F = mg, an then setting up an equation I get mdv/dt = 2mgRE.

Does this seem right. I just feel that its no correct?

 

[Doc Al]

a). Putting the origin at the Earth's surface, we get that mvo^2/2 = mgh, where h = 2RE.
Therefore rearranginf for vo we get vo = 2sqrt(gRE).
I'm skeptical whether this is right or not because this is a level 2 mechanics course, and it just seems to easy for the course and the prof.

You are right to be skeptical. The expression ΔPE = mgh is only valid near the Earth's surface, where h is small compared to the Earth's radius. Instead you have to use an expression for the gravitational PE that applies for large distances. (And for the gravitational force, use Newton's law of gravity, not F = mg.)

 

[Gogsey]

Ah I see, so does this mean the kinetic energy of the particle is equal to the integral of the Newtons law of gravation?

Where V is Vo and r is RE? And potential energy at the surface is zero since we can take this to be the origin?

 

[Doc Al]

Ah I see, so does this mean the kinetic energy of the particle is equal to the integral of the Newtons law of gravation?

Sure, but even easier to use the standard expression for gravitational PE. (See your other thread.)

Where V is Vo and r is RE? And potential energy at the surface is zero since we can take this to be the origin?

You could do it that way, but easier to just use the standard convention of setting V = 0 at r = ∞. Then find the change in PE from r = RE to r = 2RE.

 

[Gogsey]

So isn't that what I had in my original post? If not, I'm not following what you mean.

 

[Doc Al]

So isn't that what I had in my original post? If not, I'm not following what you mean.

When I said "standard expression for gravitational PE", I meant the more general one (see: http://hyperphysics.phy-astr.gsu.edu/Hbase/gpot.html#gpt") not the restricted expression that you used in your first post (mgh) which is only good near the Earth's surface.

 

[Gogsey]

So, mVo^2/2 = -GMm/r from RE to 2RE.

Maybe it was not clear but this is what I meant in my second post.

 

[Doc Al]

So, mVo^2/2 = -GMm/r from RE to 2RE.

OK, assuming you mean that ΔKE = -ΔPE as you go from r = RE to 2RE. (Just conservation of energy.)

 

[Gogsey]

For part B0, we now know Vo, we knoe RE and also we know a. Can we just use the equation for motion, d = volt -gt^2/2 and use the quadratic formula to solve for t?

 

[Doc Al]

Can we just use the equation for motion, d = volt -gt^2/2 and use the quadratic formula to solve for t?

No. That equation is only valid for constant acceleration; it doesn't apply here. The acceleration is not g, but varies with distance.