Orbital Mechanics

by Rian Cartmell, January 1998

New Quantities
Formulae
General Problem Solving Strategy
Example Problem #1
Example Problem #2
Sample Problems
Solutions to Sample Problems
Go back to Tutorial Page

Here's the relevant quantities:

```
r  distance between centers
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m₁  mass of central object
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m₂  mass of satellite
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F_g force due to gravity
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F_c centripetal force
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G  universal gravitational constant = 6.67(10^-11)Nm²/kg²

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v orbital velocity          
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T orbital period             
```

These quantities are explained on other pages.

New Quantities

Sorry! There are no new quantities here.

Formulae

Here are the relevant formulae that we have so far:

```
F_g = Gm₁m₂/r²
```
```
F_c = m₂v²/r
```

v/r = 2pi/T

And now for the only new thing here:

For the orbital condition to be satisfied, F_g = F_c


Simple substitution then gets us to the most common variant, Gm₁m₂/r²=m₂w²r.  
This simplifies to v²r = Gm₁


Note that the satellite's mass drops out here.  Substituting  2pi/T for v/r leads  to 
Kepler's observation that T²/r³ is constant for all planets in our solar system.

Go back to:  Table of Contents

General Problem Solving Strategy:

Read the problem.
Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.
Set up the orbital condition. If necessary, replace v/r with 2pi/T.
Solve it for what you're looking for. Replace known values with their numerical equivalents. For this type of problem, this is all you should ever need to do.

Go back to: Table of Contents

Example problem 1

A 1 gram fly is in orbit around a 5 kilogram bowling ball.
If it revolves at a distance of 1 m, what is the velocity of the fly?

Here is what you start with:

```
m₁ = 5kg
```
```
m₂ = .001kg
```
```
r = 1m
```
```
F_g = ?
```
```
F_c = ?
```
```
T = ?
```
```
v = ?
```

So you set up the orbital condition:  w²r³ = Gm₁
 

Solve it for v, and you get v = (Gm₁/r)^1/2.  


Toss in the numbers, and you should get  v = (6.67(10^-11)Nm²/kg²5kg/1)^1/2=.000 018m/s.  


This answer is absurdly small because the masses were also absurdly small.  


Go back to: Table of Contents

Example problem 2

A Vogon ship is in lunosynchronous orbit on the far side of the moon so that it can escape detection while waiting for final verification of Earth's termination orders. If the moon has a rotational period of 27.32 days and an equatorial radius of 1.738(10⁶), and a mass of 7.36(10²²)kg, how high above the Moon's surface is it orbiting?

WARNING! This problem contains two new tricks! The first is that it's a synchronous orbit. This means that it will always be above the same spot, thus the orbital period will be the same as the rotational period of the planet it's orbiting. (A sub-trick here is that the period is in hours-it needs to become seconds, as always.) The other is "Above the Moon's surface," which you have to convert to orbital radius by adding the equatorial radius. (note-some problems may expect you to know astronomical data. If that is the case, use the real figures.)
So here's what you really know:

m₁ = 7.36(10²²)kg
m₂ = ?
r = ?
F_g = ?
F_c = ?
T = 27.32days = 2,360,000s
v = ? Well, after solving the simplified orbital condition v²r = Gm₁
for r and substituting 2pi/T for w, you get r = (Gm₁T²/4pi²)^1/3.
Toss in numbers, and you get that r = (6.67(10^-11)Nm²/kg²7.36(10²²)2,360,000²/4pi ²)^1/3=88Mm.
Now, subtract the moon's radius, 1,738,000m, and you get about 86.7Mm above the surface.
Go back to: Table of Contents

Sample Problems
```
The answers to each problem follow it in parentheses.  They also link to a solution to 

the problem.  Try the problem, check your answer, and go to the solution if you do not

understand.
```
1.
A flea is in orbit 2.1 m from the center of a .545 kg baseball.
What is the period of orbit? What is the orbital velocity? (880 hours, 4.2x10^-6 m/s )/A>
2.
What velocity do you need to orbit 8.2 x 10⁶ m from the center of the earth? (6974 m/s or 15,600 mph)
3.
What velocity do you need to orbit 5.2 x 10⁵ m from the surface of the moon?
What would be your period of orbit? (1474 m/s, 9621 s)
4.
Fred the alien orbits the moon, completing an orbit every 3 days.
What is his velocity, and what is his radius of orbit? (2.03 x 10⁷ m, 492 m/s)
5.
What is the radius of a geosynchronous orbit?
(Around earth, T = 1 day - but convert it to seconds) How many earth radii is this? (4.22 x 10⁷ m, 6.6 Re)
6.
The moon completes a revolution about every 27 Earth days.
If you were in a luno-synchronous orbit,
how far from the center of the moon would you have to be? (8.8 x 10⁷m) Go back to: Table of Contents

Solutions to Sample Problems

1.
A flea is in orbit 2.1 m from the center of a .545 kg baseball.
is the period of orbit? What is the orbital velocity? Well, you know r=2.1m and m₁=.545kg. You are looking for T and v. So you set up the orbital condition: v²r= Gm₁. Substitute 2pi/T for v/r, then solve for T, and you get (r³/Gm₁)^1/2*2pi=T=3.17Ms=880 hrs to find v, use 2pi/T=v/r,so v=2pir/T=4.16um/s. (Go To Problem )
2.
What velocity do you need to orbit 8.2 x 10⁶ m from the center of the earth? Well, you know m₁=5.98(10²⁴kg, and r=8.2(10⁶)m. You are looking for v.
Set up the orbital condition, v²r=Gm₁. Solve for v, and you get (Gm₁/r)^1/2=v=6974m/s=15,600mph. (Goto Problem)
3.
What velocity do you need to orbit 5.2 x 10⁵ m from the surface of the moon?
What would be your period of orbit? Well, you know r=5.2 x 10⁵m and m₁=7.36(10²²)kg. You are looking for T and v. So you set up the orbital condition: v²r = Gm₁. Substitute 2pi/T for v/r, then solve for T, and you get (r³/Gm₁)^1/2*2pi=T=9621s. To find v, use 2pi/T=v/r,so v=2pir/T=1474m/s. (Goto Problem)
4.
Fred the alien orbits the moon, completing an orbit every 3 days.
What is his velocity, and what is his radius of orbit? Well, you know T = 3days=259,200s, and m₁=7.36(10²²)kg. You are looking for v and r. Let's find r first. So you set up the orbital condition: v²r = Gm₁.
Substitute 2pi/T for v/r, then solve for r and you get (Gm₁T²/4pi²)^1/3=r=2.03 x 10⁷ m. To find v, use 2pi/T=v/r,so v=2pir/T=492 m/s (Goto Problem)
5.
What is the radius of a geosynchronous orbit?
(Around earth, T = 1 day - but convert it to seconds) How many earth radii is this? You know that T = 1day=86,400s, and m₁=5.98(10²⁴)kg. You are looking for r. Set up the orbital condition: v²r = Gm₁.
Substitute 2pi/T for v/r, then solve for r and you get (Gm₁T²/4pi²)^1/3=r=4.22(10⁷)m.
Divide by the Earth's radius of 6378km, and you get 6.6 Re. (Goto Problem)
6.
The moon completes a revolution about every 27 Earth days.
If you were in a luno-synchronous orbit,
how far from the center of the moon would you have to be? You know T = 27days= 2,360,000 and m₁=7.36(10²²). You are looking for r. Well, after solving the simplified orbital condition v²r = Gm₁
for r and substituting 2pi/T for v/r, you get r = (Gm₁T²/4pi²)^1/3.
Toss in numbers, and you get that r = (6.67(10^-11)Nm²/kg²7.36(10²²)2,360,000²/4pi ²)^1/3=88Mm.
(Goto Problem) Go to: Problems Formulas Table of Contents Go back to Tutorial Page

Orbital Mechanics

Table of Contents

New Quantities

Formulae

General Problem Solving Strategy:

Example problem 1

Example problem 2

Sample Problems

1.

2.

3.

4.

5.

6.

Solutions to Sample Problems

1.

2.

3.

4.

5.

6.