Mass shot into orbit; find max distance from Earth

[oddjobmj]

Homework Statement

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.]

Homework Equations

E=[itex]\frac{1}{2}[/itex]m[itex]\dot{r}[/itex]²+U_eff(r)

U_eff(r)=[itex]\frac{L^2}{2mr^2}[/itex]+U(r)

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:

[itex]\frac{1}{2}[/itex]mv₀²=[itex]\frac{1}{2}[/itex]m[itex]\dot{r}[/itex]²+U_eff(r)

Any suggestions? Thank you for your help.

 

[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.

 

[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=[itex]\frac{1}{2}[/itex]mv²-[itex]\frac{GMm}{r}[/itex]

 

[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.

 

[Nickie]

I would first clarify the problem by stating the known variables and assumptions. The known variables are the mass of the projectile (m), the angle at which it is fired (α), and the initial speed (v0). The assumptions are that air resistance and Earth's rotation are neglected.

Next, I would approach the problem by using the conservation of energy and angular momentum. The total energy of the projectile is equal to the sum of its kinetic energy and its potential energy. Since the projectile is in orbit, its kinetic energy is equal to its potential energy. This can be expressed as:

E = K + U = 2K = 2Ueff

Using the equation for effective potential, we can solve for the maximum height that the projectile will reach:

\frac{L^2}{2mr_{max}^2}+U(r_{max}) = \frac{1}{2}mv_0^2

We can also use the equation for the initial speed to substitute for v0:

\frac{L^2}{2mr_{max}^2}+U(r_{max}) = \frac{GMm}{2R_{earth}}

Since we are looking for the maximum height, we can set the derivative of this equation with respect to r equal to 0:

\frac{-L^2}{mr_{max}^3}+\frac{dU(r_{max})}{dr} = 0

Solving for rmax, we get:

r_{max} = \frac{L^2}{GMm^2} = \frac{L^2}{GM^2v_0^2}

Substituting for L and v0, we get:

r_{max} = \frac{GM}{m^2} \cdot \frac{m^2r_{earth}^2}{GM^2} = \frac{r_{earth}^2}{M}

Therefore, the maximum height that the projectile will reach is equal to the radius of the Earth (rmax = rearth). We can express this as a ratio to the radius of the Earth:

\frac{r_{max}}{r_{earth}} = 1

In conclusion, the maximum distance from Earth that the projectile will reach is equal to the radius of the Earth, as expected from the conservation of energy and angular momentum in a circular orbit.