How to Launch a Satellite with Apogee 2.5 Times the Planet's Radius?

[Zatman]

Homework Statement

It is required to put a satellite of mass m into orbit with apogee 2.5 times the radius of the planet of mass M. The satellite is to be launched from the surface with speed v₀ at an angle of 30° to the local vertical.

Use conservation of energy and angular momentum to show that

v_0^2 = \frac{5GM}{4R}

assuming that the planet is spherical, not rotating and atmospheric effects can be ignored.

Homework Equations

L = mr^2\dot{\theta}

V = -\frac{GMm}{r}

where r is the distance from the centre of the planet to the satellite's position at time t.

The Attempt at a Solution

At t = 0, the angular momentum is L = mRv₀sin(30) = (1/2)mRv₀. Therefore:

\dot{\theta} = \frac{L}{mr^2} = \frac{Rv_0}{2r^2}

Total energy at time t is:

E = \frac{1}{2}mv^2 - \frac{GMm}{r}

E = \frac{1}{2}mr^2\dot{\theta}^2 - \frac{GMm}{r}

E = \frac{mR^2v_0^2}{8r^2} -\frac{GMm}{r}

I don't really know what to do with these equations. At the apogee, dr/dt = 0, so I could differentiate the E equation, but then I just get 0 = 0. Also I notice that E should be the same for all r as well as t, but clearly E(r) is not constant, so is the energy equation wrong?

Any help would be greatly appreciated.

 

[voko]

The energy equation is wrong, because you neglected the radial velocity.

 

[Zatman]

Well, don't I feel silly now.

So including the radial velocity gives a term with dr/dt. I set the energy equation equal to the initial energy at t=0 {i.e. (1/2)mv₀² - GMm/R}. Then at the apogee, r=2.5R and dr/dt = 0, solve for v_0 which gives the required result.

Thank you for pointing out my silly error.