1. ## Orbital Energy Problem

100 ton rocks are "thrown" from the moon to the earth.

(a) Calculate the location of the point, between the Earth and the Moon, where the gravitational field strength is zero. Give your answer in km, from the center of the Earth.
(b) What is the difference in PE between the surface of the Moon (at the point closest to Earth) and the “balance point” you found in part a? (HINT: be sure to take into account the change in PE due to the Moon’s gravity, and the change due to Earth’s.)
(c) At what speed must a barge leave the rail gun in order to just barely reach the “balance point”?
(d) If the rail gun can accelerate the barges at 100 m/sec2, how long does the rail gun have to be?

by the way:
mass of earth=5.97*10^24 Kg
mass of moon=7.37*10^22 Kg
radius of Moon's orbit= 384,000 Km

I got part a to be 341,333 km, but I don't know how to do b,c and d.

Thanks!

2. any help would be nice! Thanks!

3. Originally Posted by Linnus
(b) What is the difference in PE between the surface of the Moon (at the point closest to Earth) and the “balance point” you found in part a? (HINT: be sure to take into account the change in PE due to the Moon’s gravity, and the change due to Earth’s.)
I presume the question is asking about the change in GPE of one of those 100 ton rocks?
$\displaystyle \Delta U = -\frac{GMm}{r} - - \frac{GMm}{r_0}$
where M is the mass of the Earth/Moon and m is the mass of the rock, and the $\displaystyle r_0$ is the position you got in part a). You need to calculate the change in PE due to both the Earth and the Moon, so the formula is the same (with a different value for M), but the r's change.

Originally Posted by Linnus
(c) At what speed must a barge leave the rail gun in order to just barely reach the “balance point”?
Use conservation of energy. There are no non-conservative forces at work here, so the total mechanical energy is constant. Thus
$\displaystyle E - E_0 = 0$

$\displaystyle (T + U_E + U_M) - (T_0 + U_{E0} + U_{M0}) = 0$
where the Ts are kinetic energies, the $\displaystyle U_E$s are the GPEs from the Earth, and the $\displaystyle U_M$s are the GPE from the Moon.

Originally Posted by Linnus
(d) If the rail gun can accelerate the barges at 100 m/sec2, how long does the rail gun have to be?
You know the speed the "barge" (rock?) needs to leave the gun at from c). So you are starting a mass m from rest and accelerating it at a constant 100 m/s^2. What length of gun do you need? This is a kinematics, not gravitational, problem. Motion under constant acceleration and all that.

-Dan

4. hey, thanks! For part b why wouldn't it be
change in PE= PE at object of surface of moon- PE at the balance point
change in PE= ((-GM(moon)m/radius of moon)+(GM(earth)m/(distance from earth to surface of moon))-((GM(moon)m/distance from the moon to to balance point )+(GM(earth)m/distance from the earth to the balance point))

$\displaystyle \Delta U = (-\frac{{GM_m}m}{r} - - \frac{{GM_e}m}{r_d})- (-\frac{{GM_m}m}{r_0} - - \frac{{GM_e}m}{r_e})$

$\displaystyle {r_d}$= distanc from th earth to the surface of moon
$\displaystyle {r_0}$= distance from moon to the balance point
$\displaystyle {r_e}$= distance from earth to the balance point
because it is asking for the difference in the GPE of the surface of the moon compared to the balance point of a 100ton rock

Thanks again!

5. Originally Posted by Linnus
hey, thanks! For part b why wouldn't it be
change in PE= PE at object of surface of moon- PE at the balance point
change in PE= ((-GM(moon)m/radius of moon)+(GM(earth)m/(distance from earth to surface of moon))-((GM(moon)m/distance from the moon to to balance point )+(GM(earth)m/distance from the earth to the balance point))

$\displaystyle \Delta U = (-\frac{{GM_m}m}{r} - - \frac{{GM_e}m}{r_d}) -(-\frac{{GM_m}m}{r_0} - - \frac{{GM_e}m}{r_e})$

$\displaystyle {r_d}$= distanc from th earth to the surface of moon
$\displaystyle {r_0}$= distance from moon to the balance point
$\displaystyle {r_e}$= distance from earth to the balance point
because it is asking for the difference in the GPE of the surface of the moon compared to the balance point of a 100ton rock

Thanks again!
I'm not quite sure what your d and e subscripts are.

It should be the sum of the change in the potential energies for the Earth alone and the Moon alone, so
$\displaystyle \Delta U = (-\frac{{GM_E}m}{r_E} - - \frac{{GM_E}m}{r_{E0}}) + (-\frac{{GM_M}m}{r_M} - - \frac{{GM_M}m}{r_{M0}})$
where
$\displaystyle r_E$ is the distance between the center of the Earth and the surface of the Moon
$\displaystyle r_{E0}$ is the distance from the center of the Earth to the balance point
$\displaystyle r_M$ is the distance from the center of the Moon to the surface of the Moon
$\displaystyle r_{M0}$ is the distance from the center of the Moon to the balance point.

Also, FYI, the balance point is independent of the mass of the object we are considering. If you have a black hole instead of a 100 ton rock, the balance point is still the same.

-Dan

6. hey for part c can you say change in PE=1/2 of KE= 1/2mv^2
so PE=Mv^2 and solve for V?

Thanks!

7. Originally Posted by Linnus
hey for part c can you say change in PE=1/2 of KE= 1/2mv^2
so PE=Mv^2 and solve for V?

Thanks!
$\displaystyle \Delta E = \Delta T + \Delta U = 0$

You have your $\displaystyle \Delta U$ from part b) and if the projectile barely reaches the balance point then the final speed is 0 m/s. Thus
$\displaystyle 0 - \frac{1}{2}mv^2 + \Delta U = 0$

-Dan