Recent content by *Jessica*

Not really.. I don't understand the 60% part. It's 60% of what energy? The kinetic one I calculate in part 1?

Yes, you are right. I was just super tired, now I see it. And for part 2 I simply multiply 7W by the number of seconds in a year and then divide the number of joules by that number.

K2=0.5(1.45e20)(2.0)^2 + (-(G(1.45e20)(6e24))/6.4e6)-(G(1.45e20)(6e24))/3.84e8)

Since we need to find the kinetic energy when the asteroid hits the Earth, the potential energy is 0, leaving us with only the equation for kinetic energy. The two pieces will just move in the opposite directions. Do we use the conservation of momentum here?

1/2(mv^2). Thank you! Can you help with part 3, please?

Or do these potential energies cancel out each other and all we are left with is 1/2(mv^2)?

That means that the mechanical energy equals to the kinetic energy, since the equations are the same. Are the equations for potential energy correct?

Uhm no, I didn't take the difference. So my potential energy near the Earth will have the radius of the Earth in the denominator and the potential energy near the moon will have 384e6 m (or this number plus the radius of the Earth)? The conservation of energy equation is K1 + U1=K2 + U2. Why...

Oh yes! The radius is not squared in the potential energy equation. So I get E= -9.067e27 J. How do I find now the kinetic energy it released?

I find energy using this equation, right? E=1/2 mv^2 + (-(Gm_1m_2)/r^2) I am not sure about the velocity. This is the exact wording.

I am not that great in physics, but I put a lot of effort in trying to learn it. I would really appreciate any help on this problem, since I am not even sure where to start. An asteroid is about to fly past the moon (about 384,000km away) and is on a collision course with the earth...