# physics

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Upon its return voyage from a space mission, the spacecraft has a velocity of 24000km/h at point A, which is 7000km from the center of the earth. Determine the velocity of the spacecraft when it reaches point B, which is 6500km from the center of the earth. The trajectory between these points is outside the effect of the earth's atmosphere.

I solved for the velocity and got 2.63*10^10 and the real answer is 26300km/h. How am I off by such a big factor of 10?

• physics -

I don't know how you solved it.

INT force.dx from A to b is the added energy.

INT GMe*Ms)/r^2 dr from A to B
GMe*Ms (1/6.5E6 - 1/7.5E6)=
6.67E-11*6.0E24 Ms(2.05E-8)=8.2E6*Ms
now, add that energy to the original KE
ORigal KE= 1/2 m*(6.67E3)^2=2.2E7*m
new KE= 8.2E6M

M*3.0E7=1/2 M v^2
V= sqrt6E7

I didn't get the answer, check my work.

• physics -

i see where you went wrong. you have 7.5E6 and the second radius and it's supposed to be 7.0E6. yeah that's how i worked the problem. i converted my answer wrong when i went from m/s to km/h thank you for your help!

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