# An object of mass, m, orbits a star with mass, M, where M >> m. In general, the bound orbits are...

## Question:

An object of mass, m, orbits a star with mass, M, where M >> m. In general, the bound orbits are elliptical, which means the kinetic energy of the object T and the gravitational potential energy U may vary. For bound orbits, the total energy E is negative. If the orbit is circular, then T and U are both constant.Show that for circular orbits T : U : E = 1 : -2 : -1.

## Circular Motion

Circular motion describes when an object moves along a circular path due to being pulled towards some force at the center of the circle. This occurs in many scenarios such as a ball attached to a string or the planets orbiting the sun. The object is kept on its circular path because the central force is perfectly cancelled out by the objects speed around the circle. Instead of falling towards the center the object instead continually orbits around it. The forces on the object can be broken down into the central force, and the centripetal force due to its velocity. The centripetal force is calculated as:

{eq}F_c = mv^2/R {/eq}

Here {eq}F_c {/eq} is the centripetal force, v is the velocity of the object, m is the objects mass, and R is the distance between the object and central force.

## Answer and Explanation:

We start by recognizing that the gravitational potential between two bodies with masses m and M is:

{eq}U = -GMm/R {/eq}

We have chosen m to be the mass of the smaller orbiting object to keep the same convention as the question. The kinetic energy of the orbiting body is:

{eq}T = mv^2/2 {/eq}

Setting up our energy equation gives:

{eq}E = T+U = mv^2/2-GmM/R {/eq}

We wish to find the ratios between our energies for a circular orbit, so in order to do this we will look at the forces of our orbiting object. For an object in circular motion the only two forces are the central force and centripetal force. In our problem the central force would be the force of gravity {eq}F_g = GmM/R^2 {/eq} and the centripetal force would be {eq}F_c = mv^2/R {/eq}. In order for our object to stay in its circular orbit these forces must cancel and thus be equal.

{eq}GmM/R^2 = mv^2/R {/eq}

{eq}GM/R = v^2 \: \rightarrow \: v= \sqrt{GM/R} {/eq}

Now that we have our orbital velocity we can substitute this into our kinetic energy.

{eq}T = mv^2/2 = m \left(GM/R \right) /2 = GmM/2R {/eq}

Now that we have our kinetic energy we can find our total energy.

{eq}E = T+U = GmM/2R - GmM/R = -GmM/2R {/eq}

We have all our energies now so we can take the ratios.

{eq}T/U = (GmM/2R)/(-GmM/R) = (1/2)/(-1) = 1/(-2) {/eq}

{eq}T/E = (GmM/2R)/(-GmM/2R) = 1/(-1) {/eq}

{eq}U/E = (-GmM/R)/(-GmM/2R) = 1/(1/2) = 2/1 {/eq}

So we have found that the ratios of our energies are T:U = 1:-2, T:E = 1:-1, and U:E = 2:1

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from CLEP Natural Sciences: Study Guide & Test Prep

Chapter 3 / Lesson 2