# Central Force problems using radial motion equation

## Homework Statement

a satellite is in a circular orbit a distance $h$ above the surface of the earth with speed $v_0$, booster rockets are fired which double the speed of the satellite without changing the direction. Find the subsequent orbit.

## The Attempt at a Solution

Before the rocket boost, If we use the radial motion equation we can find the total energy of the system, that is given by:

$$E = \frac{1}{2}\dot{r}^2 + V + L^2/2r^2 = \frac{v_0^2}{2} - \frac{\gamma}{R_e + h}$$

Where I have assumed the presence of an attractive inverse square law.
and $L = rv_0$.

How can I continue from here? surely when the velocity increases, it isn't going to continue in a circular orbit.

After the boost, we have $$\frac{1}{2}\dot{r}^2 + V + L^2/2r^2 = \frac{1}{2}\dot{r}^2 - \frac{L^2}{2r^2} - \frac{\gamma}{r} = E$$

we can write

$$\frac{1}{2}4v_0^2 - \frac{L^2}{2r^2} - \frac{\gamma}{r} = E$$

then equating

I obtain

$$\frac{L^2}{2r^2} + \frac{\gamma}{r} = \frac{\gamma}{R_e + h} + \frac{3}{2v_0^2}$$

I don't see how the solutions to the above equation describe the orbit

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andrewkirk
Won't it escape the Earth? Since before the boost we have $\frac{v^2}{r}=\frac{GM}{r^2}$ where $M$ is the Earth's mass, we have $v=\sqrt{\frac{GM}{r}}$, the gravitational potential energy is $-\frac{GMm}{r}=-mv^2$ and the total potential energy is $-\frac{1}{2}mv^2$. If the speed is doubled, the KE goes to $2mv^2$ and the total energy goes to $mv^2$, which is positive, implying escape on a hyperbolic trajectory (ignoring the fact that it will be captured by the Sun).