Question

Space junk falls from rest toward the Earth from a height of 690 km above the Earth’s surface.
G= 6.673E-11 N.m ²/kg ²;
M_(earth)= 5.98E24kg;
R_(earth) = 6,380,000m;

You may assume that the junk encounters almost no air resistance above a height of approximately 20 km.


What is the speed of the junk as it falls to a height of 32.8 km?

Answer 1

any idea as how you may proceed with this problem?!

Answer 2

so confusing
so many heights

Answer 3

Not really.. when u do gravitation.. u always deal with such data.. :)

Answer 4

any formula?

Answer 5

\[v =\sqrt{\frac{ GM }{ R }}\]

Answer 6

M I RIGHT?

Answer 7

I was thinking that also, but isn't that for the speed required for a satellite to orbit the Earth without falling?

Answer 8

\[v =\sqrt{g[h1-h2]}\]

Answer 9

@Mashy where r u?we need help

Answer 10

vfinal2 = vinitial2 + [2g (sfinal - sinitial)]

The square of the final velocity equals the square of the initial velocity plus the product of twice the acceleration due to gravity times the difference in the altitudes or heights of the falling body from beginning to end.

Answer 11

Whats 'g' in that equation? Gravitational acceleration? doesn't this change with height?

Answer 12

yes

Answer 13

How would I go about using that equation to solve for velocity when acceleration is always changing :s

Answer 14

Hang on, I think I've found some relevant information. I think I will have to use work and energy procedures. I'll post back if I'm stuck or I get the question >.<

Answer 15

Oh nevermind no given mass.

Answer 16

What would i substitute in g using vfinal2 = vinitial2 + [2g (sfinal - sinitial)]