# Find the satellite's altitude above the earth's surface

• whitetiger
In summary: It seems likely to me that the potential energy should have been given as a negative number so that the difference would make sense. -1.49 x 10^11 J- 0= -1.49 x 10^11 J. That certainly makes more sense. With that, you have -1.49 x 10^11= -(GMeM)/r so that GMeM= 1.49 x 10^11 r. Now, instead of the rather ridiculous GMeM= 1.49 x 10^11 r^2, we have GMeM= 1.49 x 10^11 r and, putting that into GMeM/r=
whitetiger
When in orbit, a communication satellite attracts the Earth with a force of 15.1 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.49×10^11 J. Take the radius of the Earth to be r_e = 6.38×106 m.

Find the satellite's altitude above the Earth's surface.

We know that the gravitational potential energy is U = -(GMeM)/r, and it is given to be -1.49 x 10 ^11.

We also know the gravitational force to be Fg = (GMeM)/r^2

I am not sure, but can we set the two equations equal to each other and solve for r

whitetiger said:
When in orbit, a communication satellite attracts the Earth with a force of 15.1 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.49×10^11 J. Take the radius of the Earth to be r_e = 6.38×106 m.

Find the satellite's altitude above the Earth's surface.

We know that the gravitational potential energy is U = -(GMeM)/r, and it is given to be -1.49 x 10 ^11.

We also know the gravitational force to be Fg = (GMeM)/r^2

I am not sure, but can we set the two equations equal to each other and solve for r
Forgive me I have just started here. You might be on the right track but (and my physics is rusty) r would likely include the Earth's radius as r should be measured from the centre of gravity.

Why not F=mg U=mgh U=Fh h=U/F

Sorry what does this (below) mean?
""and the earth-satellite gravitational potential energy (relative to zero at infinite separation)""

i suspect this needs to be understood again as to whether the radius of Earth is included or not.

sirius0 said:
Forgive me I have just started here. You might be on the right track but (and my physics is rusty) r would likely include the Earth's radius as r should be measured from the centre of gravity.

Why not F=mg U=mgh U=Fh h=U/F

Sorry what does this (below) mean?
""and the earth-satellite gravitational potential energy (relative to zero at infinite separation)""

i suspect this needs to be understood again as to whether the radius of Earth is included or not.

Yes, I have tried F=mg U=mgh U=Fh h=U/F and substituting in 1.49×10^11 J/ 15.1 kN. But I've got the wrong answer.

Can someone comment more on this

Thank

I think we have to use the radius of the Earth to calculate for the satellite's altitude

So the above equation F = mgU = mgh U=Fh h= U/F is not correct

whitetiger said:
When in orbit, a communication satellite attracts the Earth with a force of 15.1 kN and the earth-satellite gravitational potential energy (relative to zero at infinite separation) is - 1.49×10^11 J. Take the radius of the Earth to be r_e = 6.38×106 m.

Find the satellite's altitude above the Earth's surface.

We know that the gravitational potential energy is U = -(GMeM)/r, and it is given to be -1.49 x 10 ^11.

We also know the gravitational force to be Fg = (GMeM)/r^2

I am not sure, but can we set the two equations equal to each other and solve for r
No, you can't "set the two equations equal to each other"! They measure different things and are not equal.

Have you copied the problem correctly? You have both gravitational force and potential energy equal to -1.49 x 10^11 (one in Newtons and the other in Joules). It is possible that they happen to be the same but rather peculiar!

Assuming that is correct, you can use the fact that Fg= (GMeM)/r^2= 1.49 x 10^11 to determine that GMeM= 1.49 x 10^11 r^2. Now put that value into GMeM/r= 1.49 x 10^11 and solve for r. Hmm, you seem to have, then,
1.49 x10^11r^2/r= 1.49 x 10^11 from which we determine that r= 1 m. That satellite is "orbiting" one meter more than 10^6 m deep in the earth! Check the problem again!

## 1. What is the formula for finding a satellite's altitude above the earth's surface?

The formula for finding a satellite's altitude above the earth's surface is: altitude = (radius of orbit + earth's radius) - earth's radius.

## 2. How do you determine the radius of a satellite's orbit?

The radius of a satellite's orbit can be determined by using the satellite's speed, the gravitational constant, and the mass of the earth. The formula is: radius = (gravitational constant * mass of earth) / (satellite's speed)^2.

## 3. Can the altitude of a satellite change over time?

Yes, the altitude of a satellite can change over time due to factors such as atmospheric drag, gravitational pull from other objects, and changes in the satellite's orbit. This is why regular adjustments and maintenance are necessary for satellites.

## 4. What is the average altitude of a satellite?

The average altitude of a satellite is approximately 22,236 miles (35,786 kilometers) above the earth's surface. This distance is known as geostationary orbit and is where most communication and weather satellites are placed.

## 5. How do satellites maintain a stable altitude above the earth's surface?

Satellites maintain a stable altitude above the earth's surface through a combination of their speed and the gravitational pull from the earth. They are also equipped with thrusters and gyroscopes that help them make small adjustments to their orbit when needed.

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