# Homework Help: Acceleration given Earth's radius

1. Dec 4, 2012

### joel amos

Given that the Earth has a radius of 6.4x10^6 m, what is the acceleration of a Trojan Badger launched from a catapult when it is 8.5x10^8 m above the surface of the Earth?

2. Dec 4, 2012

### Staff: Mentor

I believe it's just a gravity type question. Remember that a projectile's horizontal motion is at a constant velocity, neglecting air resistance. And the vertical motion is influenced by the acceleration of gravity.

What is the graviational equation that relates the attractive force between two masses and the separation distance? You should be able to use that to figure out the gravitational acceleration at that altitude. Please show your work.

3. Dec 4, 2012

### joel amos

The formula I know of is Fg = (G x m1 x m2)/ r^2 . However, I don't have the mass of the badger.

4. Dec 4, 2012

### joel amos

Here's what I've tried.
acceleration of gravity = [Radius of earth/Distance between objects]^2 x 9.8m/s/s

g = [6.4 x 10^6 m / 8.5 x 10^8 m]^2 x 9.8m/s/s
g = 5.56 x 10^-4 m/s/s

Is this correct?

5. Dec 4, 2012

### haruspex

Distance between objects? What does that mean in this context?

6. Dec 4, 2012

### joel amos

Distance between the Trojan Badger and Earth. In other words, distance from earth.

7. Dec 4, 2012

### haruspex

Think again. What answer would that give if it were 1mm from the surface of the earth?

8. Dec 4, 2012

### joel amos

Yeah, yeah, 4 x 10 ^ 20 m/s/s
I'm stumped. How would you solve this?

9. Dec 4, 2012

### Staff: Mentor

You don't need the badger's mass. What is the equation relating force to mass and acceleration?

And for the equation above, treat both masses as point masses for the purposes of calculating distance.....

10. Dec 4, 2012

### joel amos

F = ma

But, I don't know why I'd need to use the above equation to calculate distance...I already have distance.

11. Dec 4, 2012

### Staff: Mentor

That equation is why you don't need the mass of the badger. Combine the two equations, and be careful to think about what separation distance to use...

12. Dec 4, 2012

### joel amos

So it'll be...
ma = (G*m*m)/r^s simplified to a = (G*m)/r^2 ?
where...
a = acceleration of gravity
G = gravitational constant
m = mass of earth
r = distance from earth (8.5x10^8 m)
Is this correct?

And if the above is correct, then why would've my teacher included the radius of the earth into the problem?

Last edited: Dec 4, 2012
13. Dec 4, 2012

### Staff: Mentor

You are on the right track. And the radius of the Earth was in your OP problem statement...

14. Dec 4, 2012

### Staff: Mentor

And for the distance of separation.... think centers of mass...

15. Dec 4, 2012

### joel amos

Now I understand why the Earth's radius was in the original problem; it's because (as you hinted) the real distance is 8.5x10^8 m plus the radius of the earth which is 6.4x10^6 m.

16. Dec 4, 2012

### haruspex

Distance from which bit of earth, exactly? Go back and think about why my 1mm example gave a silly answer.

17. Dec 4, 2012

### joel amos

Is it this:
acceleration of gravity = [Radius of earth/(Distance from surface + Radius of earth)]^2 x 9.8m/s/s

18. Dec 5, 2012

### SammyS

Staff Emeritus
Yes.

19. Dec 5, 2012

### joel amos

g = [Radius of earth/(Distance from surface + Radius of earth)]^2 x 9.8m/s/s
g = (GM)/(Distance from surface + Radius of earth)^2

So from this thread, I can conclude that both of the above formulas can be used to get an answer. Thankfully, they both yield 5.47 x 10^-4 m/s/s, confirming this to be the answer.

Thanks to everyone for the help.