Energy to deliver 3000kg satellite into 300km orbit

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37 replies · 4K views
Δh = 13,8 km = 13,8 . 103 m
I suppose ...1 month = 30 days = 30*24 h = 30*24*3600 s = 2 592 000 s

vr = s/t = 13,8 . 103 / 2 592 000 = 0,005 m/s
 
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That is right.

By the way, in English it is a decimal point, commas are used to group digits in large numbers. Example: 1,234,567.89 (about 1.2 millions)
 
Yes, much better. Only one significant figure, but since this is a ballpark approximation and comparison it will suffice. Now you can compare the tangential and radial velocities.
 
Uffff :-) Thank you both for your patience - and of course help :-)
 
comparison: vt : vr = 7739/0,005 = 1 547 800 / 1...the tangential component of the velocity is about 1 500 000 times greater than the radial component.
 
Good. So what is your conclusion for part (b)?
 
Right.

One comment on (a): While the kinetic energy on the ground is negligible, rockets don't really care about energies. They care about speed. The rotation of Earth is a significant point in rocket launches.
 
yes, I undestand, thank you very much...