# With a circular orbit, when do I use 1/2mv^2

1. Jan 17, 2017

### jakeginobi

1. The problem statement, all variables and given/known data
At a circular orbit and at an elliptical orbit when do I use 1/2mv^2 instead of the kinetic equation from which I derived from F=GMm/r^2 which is Ek = GMm/2r

2. Relevant equations
F=GMm/r^2, Ek = GMm/2r,
Ek = 1/2mv^2

3. The attempt at a solution
For instance, when I tried to use the other kinetic energy equation (GMm/2r) to solve the kinetic energy of a satellite at perigee, it gave a totally different answer if I use 1/2mv^2

2. Jan 17, 2017

### PeroK

You need to quote the precise question and show your working.

3. Jan 17, 2017

### Staff: Mentor

Perhaps you're confusing the total mechanical energy which includes both kinetic and potential energy and is a constant over the whole orbit, with the kinetic energy alone at a particular location?

4. Jan 17, 2017

### Cutter Ketch

Your kinetic equation is valid only for circular orbits. It is derived by balancing the forces.

GMm/r2 = m v2 / r

The problem is that the r on the left is the distance from the sun (or whatever) and the r on the right is the radius of curvature of the motion. These are only the same if the orbit is circular.

5. Jan 18, 2017

### jakeginobi

In
Is total energy conserved in a circular orbit or just an elliptical orbit?

6. Jan 18, 2017

### PeroK

A circle is an ellipse with eccentricity 0.

7. Jan 18, 2017

### Cutter Ketch

In this system you have only gravitational potential energy and kinetic energy, so between those two quantities energy must be conserved for any orbit.