- #1
Gogsey
- 160
- 0
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?
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?