Solve Satellite Velocity Homework: Find Moon's Velocity

[physicaled]

Homework Statement

On the surface of a distant planet, objects experience an acceleration due to gravity of 20 m/s². The distance from the center of the planet to its surface is 8.0 * 10⁶ m. Find the velocity of a moon that is circular orbit about the planet at a height of 300,000 m above the planet's surface.

Homework Equations

a = v²/r

The Attempt at a Solution

[/B]
20 = v²/(8e6 + 300,000) = 12,884.1

I know this isn't right because my teacher said the acceleration due to gravity should not be 20. But I don't know how to solve it another way. Could anyone shed some insight?

 

[Bandersnatch]

The acceleration is caused by the gravity of the planet. Can you write the equation for gravitational acceleration at the specified height?

 

[physicaled]

The acceleration is caused by the gravity of the planet. Can you write the equation for gravitational acceleration at the specified height?

The only formula I can think to use would be a = v²/r which would leave me with a = v²/ (8e6 + 300,000) and then I'm still completely lost

 

[Bandersnatch]

What equations were introduced when you talked about gravity in class? Look in your notes, or use the internet. Here's a hint: it's all Newton's idea.

 

[physicaled]

I suppose you mean the F = GMm/r² equation? My teacher doesn't really teach so it's mostly us students trying to guess at things. So it would become:
GM/r = v² leaving me with (6.67e-11)(M)/(8e6+300,000) = v²?

I'm sorry thank you for your patience

 

[physicaled]

Actually could you solve it so that:

a= v²/r
20= v²/8e6
v²= 160,000,000
v= 12,649.1 m/s

GM/r= v²
(6.67e-11)(M)/8e6 = 12649.1²
6.67e-11M= 1.28e15
M = 1.92e25

GM/r = v²
(6.67e-11)(1.92e25)/(8e6+300,000)= v²
154293975.9 = v²
12421.5 m/s = v

Was this what you were getting at? I was staring at it a little more and the idea came to mind but I'm not really sure if it works out.

 

[Bandersnatch]

I'm not really sure if it works out.

That's pretty much what I had in mind, yes. :)
Looks about right. I'll go over the numbers later, though, I have to go now. But should be fine.

 

[physicaled]

That's pretty much what I had in mind, yes. :)
Looks about right. I'll go over the numbers later, though, I have to go now. But should be fine.

No worries! Thank you so much for your help!

 

[Bandersnatch]

Yup, numbers look good too.
Myself, I'd probably write $ma=\frac{GMm}{R^2}$, get the mass of the planet from there, and then plug it into $V=\sqrt{\frac{GM}{R+r}}$. Saves you an equation, and is more straightforward reasoning-wise.

Do you understand why you used those equations?