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

In summary, the problem involves a satellite in orbit just above the surface of a spherical planet with the same radius and acceleration as Earth. The task is to calculate the speed and time for one complete orbit. Using the equations for speed, centripetal acceleration, and centripetal force, we can calculate the speed and time by taking into account the radius of the planet and the acceleration due to gravity. The final solution is a speed of 7.9e3 ms-1 and a time of 85 minutes for one orbit.
  • #1
mogsplanet
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Homework Statement


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

Homework Equations


Speed (v) = (2pi)r/T

Cent. Accel. (a) = v2/r

Cent. Force. F = mv2/r = mw2r

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 into v2/r but this is heading down a blind alley.
Please Help.
 
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  • #2
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:
[tex]F= -\frac{GmM}{r^2}[/tex]
"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
[tex]a= -\frac{GM}{r^2}[/tex]

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 [itex]r^2= 4(6400)^2[/itex].
 
  • #3
hi mogsplanet! welcome to pf! :smile:

(try using the X2 icon just above the Reply box :wink:)
mogsplanet said:
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 into v2/r …

what does the year have to do with it? :confused:

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? :wink:
 
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  • #4
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
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
 

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

1. What is centripetal acceleration?

Centripetal acceleration is the acceleration that an object experiences as it moves in a circular path. It is always directed towards the center of the circle and its magnitude is equal to the square of the object's speed divided by the radius of the circle.

2. How is centripetal acceleration related to a satellite orbiting above Earth?

In the context of a satellite orbiting above Earth, centripetal acceleration is responsible for keeping the satellite in its circular path around the planet. It is caused by the gravitational force exerted by Earth on the satellite.

3. How is the centripetal acceleration of a satellite above Earth calculated?

The centripetal acceleration of a satellite above Earth can be calculated using the formula a = v^2/r, where "a" is the centripetal acceleration, "v" is the orbital speed of the satellite, and "r" is the radius of the satellite's orbit.

4. Does the Earth's radius affect the centripetal acceleration of a satellite?

Yes, the Earth's radius does affect the centripetal acceleration of a satellite in orbit. As the radius increases, the centripetal acceleration decreases, meaning the satellite's speed must also decrease to maintain its orbit.

5. How does the centripetal acceleration of a satellite change as it orbits Earth?

The centripetal acceleration of a satellite remains constant as it orbits Earth, assuming the distance from Earth remains constant. However, if the satellite's distance from Earth changes, the centripetal acceleration will also change to maintain the satellite's orbit.

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