# Centripetal Acceleration of a satellite above a planet of the Earth's radius.

1. Jul 6, 2011

### mogsplanet

1. The problem statement, all variables and given/known data
A satellite is in orbit just above the surface of a spherical planet which has the same radius as Earth and the same acceleration of free fall at it's surface. Calculate:
i) Speed
ii) Time for 1 complete orbit.

Radius of Earth = 6400km or 6400000m and accel. of free fall = 9.81ms-2
2. Relevant equations
Speed (v) = (2pi)r/T

Cent. Accel. (a) = v2/r

Cent. Force. F = mv2/r = mw2r

3. The attempt at a solution

First time on here so finding it difficult to locate symbols so sorry for loss in translation.

My attempt of a solution is: v = (2xpi)x 6400000/3.1e7 (no of s in 1 year (365.25days))

v = 1.29ms-1
then plugged this in to v2/r but this is heading down a blind alley.
Please Help.

2. Jul 6, 2011

### HallsofIvy

If you can't use LaTeX, "^" is standard for powers. Or [ sup]2[/sup] (without the first space).

Why would you believe "v = (2xpi)x 6400000/3.1e7" Are you assuming the satellite takes one year to orbit the planet???

The simplest thing to do is to calculate the centripetal force on the satellite- that is just the force holding in orbit which is just the gravitational force on the satellite:
$$F= -\frac{GmM}{r^2}$$
"G" and "M" are those values for the earth and m is the mass of the satellite, r is 6400 km.

And then, of course, F= ma so
$$a= -\frac{GM}{r^2}$$

Now, on the surface of the earth, the acceleration due to gravity is -9.81 and the radius is, to two significant figures, 6400 km! Adding another 6400 for the altitude of the satellite, we have r= 2(6400) so $r^2= 4(6400)^2$.

3. Jul 6, 2011

### tiny-tim

hi mogsplanet! welcome to pf!

(try using the X2 icon just above the Reply box )
what does the year have to do with it?

as usual, use good ol' https://www.physicsforums.com/library.php?do=view_item&itemid=26" F = ma …

what is the acceleration? what is the force?

Last edited by a moderator: Apr 26, 2017
4. Jul 6, 2011

### mogsplanet

The question has no mention of mass so this should not be used. All the information I have is in the original question. It is A satellite in low orbit around a planet with same radius as the Earth.

5. Jul 6, 2011

### mogsplanet

Thanks for help. It has come to me now.

9.81 = v^2/6400000

So v = 7.9e3 ms-1

and so time for 1 orbit is (2pi)r/v

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