Question

The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were observed by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit--a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass M.

For all parts of this problem, where appropriate, use G for the universal gravitational constant.


Part A) Find the orbital speed v for a satellite in a circular orbit of radius R. (Express the orbital speed in terms of G, M, and R),

Part B) Find the kinetric energy K of a satellite with mass m in a circular orbit with radius R. (Express your answer in terms of m, M, G, and R).

Part D) Find the orbital period T. (Express your answer in terms of G, M, R, and pi).

Part F) Find L, the magnitude of the angular momentum of the satellite with respect to the center of the planet. (Express your answer in terms of m, M, G, and R.

Part G) The quantities v, K, U, and L all represent physical quantities characterizeing the orbit that depend on radius R. Indicate the exponent (power) of the radial dependence of the absolute value of each. (Express your answer as a comma-separated list of exponents corresponding to v, K, U, and L, in that order.

Answer 1

A) Orbital speed: $v=\sqrt{\frac{GM}{R}}$

B) Kinetic energy: $K= \frac{GmM}{2R}$

D) The orbital period is $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F) The angular momentum is $L=m\sqrt{GMR}$

G) Exponent of radial dependence:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Explanation:

A)

We know that for a satellite in circular orbit around a planet of mass M, the gravitational force between the satellite and the planet is

$F=G\frac{mM}{R^2}$

where m is the mass of the satellite.

This force provides the centripetal force needed for the circular motion, which is

$F=m\frac{v^2}{R}$

where v is the orbital speed.

Since the gravitational force provides the centripetal force, we can equate the two expressions:

$G\frac{mM}{R^2}=m\frac{v^2}{r}$

And solving for v, we find

$v=\sqrt{\frac{GM}{R}}$

B)

The kinetic energy of an object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this problem,

m is the mass of the satellite

$v=\sqrt{\frac{GM}{R}}$ is the speed of the satellite (found in part A)

Substituting, we find an expression for the kinetic energy of the satellite:

$K=\frac{1}{2}m(\sqrt{\frac{GM}{R}})^2 = \frac{GmM}{2R}$

D)

The orbital speed of the satellite can be rewritten as the ratio between the distance covered during one orbit (the circumference of the orbit) divided by the period of revolution:

$v=\frac{2\pi R}{T}$

where

$2\pi R$ is the circumference of the orbit

T is the orbital period

We already found that the orbital speed is

$v=\sqrt{\frac{GM}{R}}$

Substituting into the equation,

$\sqrt{\frac{GM}{R}}=\frac{2\pi R}{T}$

And making T the subject,

$T=\frac{2\pi R}{\sqrt{\frac{GM}{R}}}=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

F)

The angular momentum of an object is defined as

$L=mvr$

where

m is the mass of the object

v is its speed

r is the radius of the orbit

For the satellite here we have

m (mass of the satellite)

$v=\sqrt{\frac{GM}{R}}$ (orbital speed)

R (orbital radius)

Substituting,

$L=m\sqrt{\frac{GM}{R}}R=m\sqrt{GMR}$

G)

First, we rewrite the list of expressions for the different quantities that we found:

Orbital speed: $v=\sqrt{\frac{GM}{R}}$

Kinetic energy: $K= \frac{GmM}{2R}$

Orbital period: $T=\frac{2\pi}{\sqrt{GM}}R^{3/2}$

Angular momentum: $L=m\sqrt{GMR}$

Now we observed the dependence of each quantity from R:

Orbital speed: $v\propto R^{-1/2}$

Kinetic energy: $K \propto R^{-1}$

Orbital period: $T \propto R^{3/2}$

Angular momentum: $L \propto R^{1/2}$

So the exponent of the radial dependence of each quantity is:

Speed: -1/2

Kinetic energy: -1

Orbital period: 3/2

Angular momentum: 1/2

Learn more about circular motion:

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