# Calculus in the velocity and acceleration of satellites.

Tags:
1. Jul 16, 2017

### VinnyO

1. The problem statement, all variables and given/known data
I am working on a project dealing with the velocity and acceleration of satellites based on their distance from Earth. I was recommended to include some calculus in this project. Originally I thought I could just take the derivative of the orbital speed equation to find acceleration, given the mass of the Earth and the gravitational constant would be constants in the situation. When I tested this hypothesis with the accepted acceleration equation the answers came out different. I am fairly sure this is because the remaining variable is R (the radius of the orbit) as opposed to time, but perhaps I am making a different mistake. So at this point I am in search for a way to make calculus relevant in this situation.

2. Relevant equations
v = SQRT(G*M/R)
a = G*M/R^2

3. The attempt at a solution
I also have been looking for a way to calculate the distance of objects like the ISS from Earth based on observations/ measurements I can take myself (from Earth), but I assume any solution to this would be related to trigonometry or geometry.

2. Jul 16, 2017

### haruspex

Remember that velocity and acceleration are vectors. Differentiating a speed (a scalar) does not give acceleration.

3. Jul 16, 2017

### Kaura

Remember that acceleration is change in velocity with respect to change in time

4. Jul 16, 2017

### Staff: Mentor

Moved to Calculus & Beyond section

5. Jul 16, 2017

### VinnyO

Currently I have only taken one year of calculus in which I have not learned vectors (the report is for math in the program I am in and because I have taken calculus I was encouraged to utilize that skill). However with a bit of research on the basics of such I realize how obvious my mistake is. With that in mind my first though was that perhaps I could find velocity given the acceleration formula with some integration. But anyways in short this was my math:
I took an online example where R (the radius of the orbit) was 6.47 x 10^6 m. When plugged into the acceleration formula, a = (6.673 x 10^-11 N m^2/kg^2) • (5.98 x 10^24 kg) / (6.47 x 106 m)^2, a (acceleration) was equal to 9.53 meters per second squared.
To test my idea to use calculus I took the orbital speed equation, v = SQRT(G*M/R), and plugged in the gravitational constant, 6.673 x 10-11 N m^2/kg^2, for G and the mass of Earth, 5.98 x 10^24 kg, for M
I than simplified this to v=998806037 / 50*R^1/2
The derivative of this was a= -998806037 / 100R^3/2
I finally just plugged in the same R value into the new potential acceleration equation, a= -998806037 / 100(6.47x10^6)^3/2, and got a=6.07x10^-4 which was not equal to the answer produced by the equation so I knew I made some wrong assumption.

6. Jul 16, 2017

### haruspex

That is the derivative wrt radius, dv/dR. Acceleration is the derivative of velocity wrt time, dv/dt.