Recent content by teh_game

is this it? http://en.wikipedia.org/math/8923c2d8315043f3ef49ca7169c402ec.png nope, it's this: http://en.wikipedia.org/math/a568f1657e033b789d0bc19068487ffc.png but I'm not sure how to do that little 3/2 thing on the side...

ummm, what's the equation?

So, whenever the Earth is in the middle, Mars is brighter, regardless of the model. Is that correct? *need some reassurance* :) thanks a lot ! Edit: wait, the question already tells us that :bugeye: I know that for the copernican model, it's because the distance between Mars...

Oh, great! Thanks! :smile:

oh okay, thanks for the tip! (I created a new thread for something else :smile:) anyways, the chair he's sitting on is pushing him up? actually, i don't think anything really is pushing him up since the space shuttle is, as you said, falling as he is. Therefore, he is experiencing...

Hey, I've got a question which I hope you guys can help me with. It is observed that Mars is somewhat brighter when in opposition (ie. sun- Earth - Mars in that order) that at other times. How can this be explained using a) The Ptolemaic model and b) The Copernican model of the solar...

Okays, this is another question... I believe I answered correctly Critically assess the following statement: `Astronauts in a space capsule orbiting the Earth at a height of 900 km experience weightlessness: hence the gravitational field due to the Earth must be zero at that altitude...

Great! thanks!

I have a question, can you calculate the gravitational field with the formula: g = GM/d^2 ? I'm getting confused because one website included a negative sign and the other didn't!

7.903 \times 10^3\ m/s \ is\ the\ orbital\ velocity\ that\ needs\ to\ be\ given. yay!

Oops, I made a mistake... I thought G was 9.8, but it's actually the "gravitational constant" which is 6.668*10^-11 Edit: hehe that's why my V was over the speed of light ! :smile:

oh, okays... my bad :blushing: but I'm pretty sure i got it now! From the equation: mv^2/6.378*10^6 = 9.8*5.974*10^24*m/(6.378*10^6)^2*m (cancel the m(object)) v = 3,029,726,249 And if the mass was three times as much, it wouldn't matter! Thanks! (ummm, warn me if I'm...

The orbital velocity is: 29.4 (9.8 * 3) I think I'm right, but I'm not 100% sure that v = 9.8

v = 9.8 Therefore v^2/r = G*m/r^2 9.8^2/r = 9.8mr^2 m=9.8/r Therefore if the other satellite is 3 times m then 3m = 9.8*3 / r ummm, does that even make sense ? :confused:

okay, I have another question: 10. A science group put in a satellite of mass m kg into a circular Earth orbit of radius r. The orbital velocity it needs to remain in this orbit is v. They now put another satellite into a similar orbit at the same altitude. Its mass is 3 times m. What orbital...