# Mass shot into orbit; find max distance from Earth

1. Nov 14, 2013

### oddjobmj

1. The problem statement, all variables and given/known data
A projectile of mass m is fired from the surface of the earth at an angle α from the vertical, where α = 0.46 radians. The initial speed v0 is equal to √{GM/Rearth}. Calculate the maximum height that the projectile will reach. Neglect air resistance and the earth's rotation. Express the result as the ratio to the radius of the Earth. [Hint: Apply the conservation laws to the orbit of the projectile.]

2. Relevant equations

E=$\frac{1}{2}$m$\dot{r}$2+Ueff(r)

Ueff(r)=$\frac{L^2}{2mr^2}$+U(r)

3. The attempt at a solution

Intuitively speaking the mass will go into orbit and because we're not necessarily near the surface of the earth the potential is not mgh. Also, the orbit must be bound or the answer would be infinity. If the mass is in orbit I could compare the effective potential of two points in the orbit but I don't see how to make a useful comparison with the information I am provided. What would be the variable of comparison?

Should I be comparing the initial kinetic energy to that of a mass in orbit? I can't seem to figure out how to cancel out v02 if that is the case:

$\frac{1}{2}$mv02=$\frac{1}{2}$m$\dot{r}$2+Ueff(r)

Any suggestions? Thank you for your help.

2. Nov 14, 2013

### rcgldr

If the object is fired from the surface at some angle other than zero, then eventually the object will impact back into the earth, unless it reaches or exceeds escape velocity.

Getting back to the problem statement, you're given the initial energy (potential + kinetic) and the angle. You need to determine when the potential energy is at a maximum and kinetic energy is at a minimum, which will correspond to the maximum height. You may need to determine the parameters of the elliptical path.

3. Nov 14, 2013

### oddjobmj

I was given the speed. I don't have the object's mass. I don't know how to represent its energy in terms of v0 besides 1/2mv^2 which won't work here.

The potential is at a maximum at the apogee. How I know where the apogee is I am not sure.

E=$\frac{1}{2}$mv2-$\frac{GMm}{r}$

Last edited: Nov 14, 2013
4. Nov 14, 2013

### haruspex

Try comparing energy (KE+PE) and angular momentum about Earth's centre between launch and apogee. You won't need to know the mass - that will cancel out.