# A satellite in orbit fires it's engines for a short interval. Is the new orbit closer or further away?

A satellite is in a circular orbit when its engines turn on to exert a small force in the direction of the velocity for a short time interval. Is the new orbit further or closer to the Earth?

The solution is that the new orbit is further away (which is also intuitive) and is justified by stating that there is a positive increase in the total energy which is given by the formula: $$\:E_{T}=-\frac{GMm}{2r}$$ And that states that an increase in the total energy would result in a larger radius. Thus, the new orbit is further away.

However, the problem arises when I look at this equation which relates the speed with the radius:

$$\:v^2=\frac{GM}{r}$$ Since the small force was in the direction of the velocity, an increase in veloctiy should result in a DECREASE in the radius meaning that the new orbit is closer in according to this equation.

Why is it that the first equation is correct to use, while the second one is wrong? Why is the second equation not working here?

Thanks!

• $E_T$ is only the potential energy; you're ignoring the kinetic energy. – Dave Oct 28 '13 at 21:07
• @Dave: Actually, that formula takes into account both potential, and kinetic energy. You can see how it is derived here link – Sheheryar Zaidi Oct 28 '13 at 21:10
• Oh yeah; I missed the factor of (1/2) – Dave Oct 28 '13 at 21:13

Others correctly state that the trajectory is no longer circular. Your second equation $$v^2=\frac{GMm}{r}$$ was probably derived from uniform circular motion: $$\frac{mv^2}{r}=\frac{GMm}{r^2}.$$
As for your first equation, the more general form is $$E_T=-\frac{GMm}{2a},$$ where $a$ is the semi-major axis. For circular orbits, $a=r$ (the radius of the orbit). Your previous reasoning still works here: the total energy of this two-body system arguably increases, $a$ must increase as well since an increase in $a$ would correspond to an energy closer to zero; that is, higher than before.
• @sheheryar-zaidi: Satellite has gravitational acceleration $g$ at each point of its trajectory. Orbit can be a circle when $\frac{v^2}{r}$ (inward acceleration of circular motion) equals $g$. – user1086737 Oct 29 '13 at 16:38