Recent content by bmoore509

T = 2pi (sqrt((R^3)/GM)) R3= (4190000+14120300)3 = 6.138840926x1021 G*M = (6.67259x10-11)*(1.710200237x1024)=1.1411465x1014 sqrt ( R3/(G*M) = sqrt (6.138840926x1021/1.1411465x1014) = sqrt (53795379.7) = 7334.533366 T=2pi(sqrt(r^3/GM) = 2*pi*7334.533366 = 46084.23228 Does that...

Oh! Thank you! Let me rework this then.

I redid my work using the formula as you wrote it. T = 2pi (sqrt((R^3)/GM)) R3= (4190000+14120300)3 = 6.138840926x1021 G*M = (6.6725x10-11)*(3.265952291x1025)=2.17923606x1015 sqrt ( R3/(G*M) = sqrt (6.138840926x1021/2.17923606x1015) = sqrt (281696.23) = 1678.382921...

Homework Statement Given: G = 6.67259 × 10−11 Nm2/kg2 The acceleration of gravity on the surface of a planet of radius R = 4190 km is 6.5 m/s2. What is the period T of a satellite in circu- lar orbit h = 14120.3 km above the surface? Answer in units of s. Homework Equations...

Can anyone try to help me? Thanks.

But how do I figure out how to draw Lw?

Sorry. I'm just having a really hard time visualizing this. I'm really not sure what I'm supposed to be doing here. :confused:

http://upload.wikimedia.org/wikipedia/commons/0/09/Torque_animation.gif So I looked at that trying to understand. (Just want you to know where I get my logic from. Haha) So momentum is the same direction as Torque but the angular momentum, you can do sort of the right hand rule. I think I...

So if that's the rate of change of the angular momentum, how does that tell us the direction of the angular momentum?

I'm having a total brain freeze. The other formula for torque is what?

So then it'd be pointed to the left of the axle. Ugh. I'm no good at this.

Well, my original thought was that it's the opposite side of the axle. Like the direct opposite.

Oh my. Haha. I meant pointing away from us.

I got it! You were right. So I used wolfram online to calculate it.

So Torque would actually be pointing away then. If I'm picturing everything correctly.