# Work done to change circular orbit and orbital speed

If a satellite of mass $$m$$ is orbiting a planet of mass $$M$$ with radius $$r_1$$ and orbital speed $$v_1$$ and is brought to orbit at $$r_2$$ with speed $$v_2$$, its kinetic energy changes by a quantity

$$\Delta E_k = \frac{1}{2} GMm \left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$

Since the total energy in orbit 2 is equal to the total energy in orbit 1 plus the work done to change orbit:

$$E_{tot}^{(2)}=E_{tot}^{(1)}+W_{12}$$ $$\frac{1}{2}mv_1^2 - \frac{GMm}{r_1} = \frac{1}{2}mv_2^2 - \frac{GMm}{r_2} + W_{12}$$

I obtain that the work done to change orbit is

$$W_{12}= \Delta E_k - \Delta E_p = -\frac{1}{2} GMm \left(\frac{1}{r_2}-\frac{1}{r_1}\right)$$

I thought I could calculate this assuming that the minimum work required to change orbit is that done along a radius (either against or with gravity) by a force equal to gravity (in magnitude). Using the definition of work done:

$$W_{12} = \intop_{r_1}^{r_2} \frac{GMm}{r^2} dr$$

I don't see how to obtain the factor $$1/2$$. What am I doing wrong?

• The problem is that you're using the gravitational force to calculate the work done to change the orbit. That will never happen. The work done is exerted by a non conservative force from the engines. Oct 14, 2018 at 18:49
• Wouldn't a force that is equal and opposite gravoty be enough to do that work though? Oct 14, 2018 at 19:17
• No, it also changes KE. Oct 14, 2018 at 21:21

The framing of the problem makes it sound like a satellite is switching from one circular orbit to another.

You have the force due to gravity in the last integral, not the force due to whatever is changing the orbit. The central gravity alone will not change the object's orbit, so there must be another force that is doing it. When computing changes in total energy you must take in to account the external forces acting on the system.

• Should the work done by this external force not be equal to the work done by gravity? Oct 14, 2018 at 18:46

I thought I could calculate this assuming that the minimum work required to change orbit is that done along a radius (either against or with gravity) by a force equal to gravity (in magnitude).

But the work done by that force can only change the gravitational potential energy and not the kinetic energy, correct?

Put another way, if there is just an external force applied to the object that is equal in magnitude and opposite in direction to the gravitational force, then the object has zero net force and so there will be no change at all in the kinetic energy of the object by such a force.

But, to change from one circular orbit to another (in the same plane) requires that the kinetic energy change by (negative) 1/2 the change in gravitational potential energy.

Thus, the work done by such a force cannot be equal to the work done in changing to a different circular orbit.