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Central Force problems using radial motion equation

  • #1
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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.

Homework Equations


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
 

Answers and Replies

  • #2
andrewkirk
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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).

So I guess by orbit, they mean the hyperbolic path it will follow. The equations for that path should be readily obtainable from a page on orbital trajectories such as this.
 

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