# Determining speed of orbiting object

1. Jan 28, 2015

### brake4country

1. The problem statement, all variables and given/known data
The earth has a radius of approximately 6400 km. If an object could orbit the earth just at its surface, how fast would it have to travel?

2. Relevant equations
C = 2πr; ac=v2/r; Fc=mv2/r

3. The attempt at a solution
Circumference = 12800π. I tried to solve for the rate by d = r x t (86400s = 24 hours) but my answer is very small.

2. Jan 28, 2015

### Nathanael

So I suppose you used the time it takes the Earth to rotate (the time of 1 day). If so then what you calculated is actually the speed with which WE are moving relative to the center of Earth right now. But there's no reason an orbiting object should take the same time to complete one cycle. (If it did happen to take the same time, then we would feel weightless!)

This is the force that is required in order to travel in a circle of radius r at a speed of v
In the case of an object orbiting, what is the source of the centripetal force?

3. Jan 28, 2015

### brake4country

The source of the centripetal force is the gravitational force from the earth. But no mass was given!

4. Jan 28, 2015

### Nathanael

Hmm... Does gravitational acceleration depend on mass?

5. Jan 28, 2015

### brake4country

Oh, I see: Fg=Fc
GMm/r2 = mv2/r
GM/r = v2
(6.67x10^-11)(6x10^24)/(6.4x10^6) = 7907.6 m/s or 8 km/s

6. Jan 28, 2015

### brake4country

To answer your question, gravitational acceleration of objects does not depend on mass but I needed the mass of the Earth to calculate this one. The problem did not give it. Am I to know the mass of the earth?

7. Jan 28, 2015

### brake4country

Hmmm. I just realized that I could have just used mg = mv2/r since the orbital is very close to the surface. Blah!

8. Jan 28, 2015

### Nathanael

Right. :)

I had the same thought, "if they tell you the radius of Earth, they ought to tell you the mass also!" but then I soon realized that you can just use g=9.8

9. Jan 28, 2015

### dean barry

Treat the orbiting mass as negligible (which it is, compared to the earth)
Have a look at the attached sheet.

File size:
66.5 KB
Views:
95