# Force Magnitudes, directions, gravity, etc (lost)

• PianoMan

#### PianoMan

Hey all,

First-time poster here, hope y'all can help me. I'm ok with 85% of my homework this evening, but this one darned problem keeps tripping me up.

Real quick, just to note, I'm still in high school, but this is Physics AP, so let me know if I'm in the wrong forum. Thanks!

(a) Calculate the magnitude of the gravitational force exerted on a 425-kg satellite that is a distance of two Earth radii from the center of the earth.

(b) What is the magnitude of the gravitational force exerted on the Earth by the satellite?

(c) Determine the magnitude of the satellite's acceleration.

(d) What is the magnitude of the Earth's acceleration?
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Now, I know somehow I need to incorporate this formula:

W = G * (Me m / r^2)

W is the weight. I don't have this because I have a mass but not an acceleration.

G is always the same (6.67 * 10^-11)

Me (mass of earth) = 5.98 * 10^24kg

m (mass) = 425 kg

r (radius of earth) = 6.38 *10^6

So I know I can just plug and chug in that formula and solve for W. But I don't think that really helps me.

Am I way off here? Or just missing a small detail? And now I'm also reading that as the distance increases, the gravitational force decreases. How in the world do I incorporate this?

EDIT- For what it's worth, I went ahead and solved for W. I got 3.45 * 10^30

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Originally posted by PianoMan
Hey all,

First-time poster here,

Welcome to the forums!

Real quick, just to note, I'm still in high school, but this is Physics AP, so let me know if I'm in the wrong forum. Thanks!

This forum is right.

W = G * (Me m / r^2)

W is the weight. I don't have this because I have a mass but not an acceleration.

W is what you're solving for with the equation for part 1. Weight is the gravitational force. 9.8m/s^2 is the gravitational acceleration only on the surface of the earth.

Do this when you're finished with this problem, just to prove it to yourself. Find the acceleration of gravity on the surface of the Earth. r=6378km. You'll see that mass of the smaller body cancels out.

EDIT- For what it's worth, I went ahead and solved for W. I got 3.45 * 10^30

You'd better check that again. Things in orbit weigh less than they would on the surface. Make sure you've got ^(-) and ( ) in the correct places.

Also, you always always always want to keep your units in the problem. Remember when we lost the Mars probe a few years back? It's cause the NASA guys left them off, assuming the other folks knew what they were talking about.

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So I know I can just plug and chug in that formula and solve for W. But I don't think that really helps me.

Am I way off here? Or just missing a small detail?

Yes, you are missing a small detail: solving for W helps you because "magnitude of the gravitational force" IS weight! That's what you are asked for.

And now I'm also reading that as the distance increases, the gravitational force decreases. How in the world do I incorporate this?

You've already incorporated it- "r" in the denominator of your formula is the distance. As the distance increases, the denominator of the fraction increases, the fraction itself decreases.

I have no idea how you got "10^30" in the weight calculation:
The numbers involved are m= 4.25x10^2, Me= 5.98x10^24, G= 6.67x10^(-11) and r= 1.267 x 10^7 (I doubled the radius of the Earth because the problem says "twice the radius of the earth"). Ignoring the digits, the power of 10 is 2+ 24- 11- 2(7)= 26- 25= 1. You should have some number times 10.