- Good morning.
Today, we are going to determine the altitude of a satellite in geosynchronous orbit.
Flipping Physics
Geosynchronous orbit, which is also called geostationary orbit,
is where the object orbiting the Earth is always above the same spot on the Earth.
Realize that in order for an object to stay above the same location of the Earth, the orbiting object must be
directly above the Earth’s equator.
In order to answer this question, we need some known values.
According to NASA, the mass of the Earth is 5.9723 times 10 to the 24th kilograms and the equatorial
radius of the Earth is 6.378 times 10 to the 6th meters.
Bobby, how shall we start solving for the altitude of a satellite in geostationary orbit?
- I got nothing.
- All our knowns are in base SI units, so we don’t need to convert anything.
- Draw a picture?
- There you go. I drew your picture.
- Thanks
- We need to draw a free body diagram of the forces acting on the satellite.
The only force acting on the satellite is the force of gravity caused by the Earth on the satellite
and it is pointed toward the center of mass of the Earth.
- There is also an equal but opposite force of gravity caused by the
satellite on the Earth which is pointed toward the center of mass of the satellite.
Billy, you are correct, however, because we are finding the altitude of the satellite,
we are currently only interested in the forces acting on the satellite.
And because the force from the satellite on the Earth acts on the Earth and not the satellite,
we are not going to include that force in our free body diagram.
Bo, what now?
- We've drawn a free body diagram, so let's sum of the forces.
[Mr. P] In what direction do we need to sum the forces in?
Yeah. The in-direction.
The net force in the in-direction equals the force of gravity and it equals mass times the acceleration
in the in-direction.
Which is the centripetal acceleration.
- So the net force in the in-direction is the force of gravity
and therefore is the centripetal force which keeps the satellite in orbit around the Earth.
That makes sense.
- We use the Big G equation for the force of gravity.
- That’s Newton’s Universal Law of Gravitation.
- Right. So Big G.
- The Universal Gravitational Constant.
- Yes. Big G times mass 1 times mass 2 all divided by r squared,
where r is the distance between the centers of mass of the two objects.
For centripetal acceleration we can use either radius times angular velocity squared or
tangential velocity squared divided by the radius, but I don’t know which one to use.
Well, what do we know about the satellite?
- It’s in geosynchronous orbit.
- So it revolves at the same rate the Earth rotates.
- So we can find the angular velocity of the satellite.
Angular velocity equals change in angular position over change in time.
The Earth and the satellite go through 1 revolution or 2 pi radians every 24 hours.
But we need angular velocity in terms of seconds because it needs to match the Universal Gravitational
Constant which is in newtons times meters squared divided by kilograms squared.
- And newton’s are in kilograms times meters per second squared.
- Right! So multiply that by 1 hour over 3600 seconds
to get the angular velocity of the satellite and Earth to be
7.27221 times 10 to the negative 5th radians per second.
- Great. Now that we know we are using angular velocity and not tangential velocity, let’s identify our masses.
In Newton's Universal Law of Gravitation, one of the two masses is the mass of the
satellite the other one is the mass of the Earth.
And then the satellite is the object which is in orbit. The object which has the angular velocity.
- Billy and Bobby: Every body brought mass to the … Bo: Everybody brought the mass of the satellite
to the party.
Right. There are two masses. We need to specify it is the mass of the satellite which
everybody brought to the party.
"Everybody brought mass"
Bobby, what about the two r’s in this equation. Could you please tell me what they are
specifically in this example?
- Well, the r in the universal law of gravitation
is the distance between the centers of mass of the two objects.
So in this problem that is the distance from the center of mass of the Earth to the center of mass of the satellite.
And the r which came from centripetal acceleration equals r times angular velocity squared
is the radius of the orbit of the satellite.
And those are the same thing in this problem. Because this is an example where the r in
the universal law of gravitation is the radius.
- Which means we can divide the whole equation by radius and we get radius cubed in the denominator on
the left side.
Also, the radius of the satellite equals the radius of the Earth plus the satellite’s altitude, or its vertical height
above sea level.
Let’s solve for the satellite’s orbital radius, that r, first and then we will solve for the altitude of the satellite.
First we can multiply the whole equation by radius cubed and divide by angular velocity squared.
Then we can take the cube root of the whole equation to get the
radius equals the cube root of the universal gravitational constant times
the mass of the earth divided by angular velocity squared.
Plugging in our numbers gives us a geostationary orbital radius of 4.22323 times ten to the 7th meters.
From which we can subtract the
equatorial radius of the earth to get 3.58543 times ten to the 7th or 3.59
times ten to the 7th meters with three significant digits.
And, if you visit NASA’s website, you will discover they have published the geosynchronous
geosynchronous orbital radius as
35,900 km, which is exactly what we calculated.
Which means we get to say, “The physics works.
The physics works. Uh huh uh huh. The physics works.” Bo: Does that even qualify?
We did not test our results.
- I think we can trust that NASA has tested this with real satellites.
- Billy: I think we can trust NASA. Bo: Eh.
- The physics works.
Thank you very much for learning with me today.
I enjoyed learning with you.