Calculating Energy for Satellite Orbit at 300km | Physics Homework Help

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The discussion focuses on calculating the energy required to launch a satellite into a circular orbit at 300 km above Earth's surface, with a mass of 3,000 kg. The work done is calculated using gravitational potential energy equations, resulting in approximately 9.81 x 10^10 J, while the initial kinetic energy from Earth's rotation is deemed negligible. Additionally, the radial component of velocity, resulting from a 13.8 km descent over one month, is compared to the tangential velocity, revealing that the tangential component is significantly greater, approximately 1,500,000 times larger. The conversation emphasizes the importance of unit conversions and the relevance of Earth's rotation in satellite launches. Overall, the calculations and comparisons illustrate the dynamics of satellite motion and energy considerations in orbital mechanics.
  • #31
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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  • #32
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)
 
  • #33
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.
 
  • #34
Uffff :-) Thank you both for your patience - and of course help :-)
 
  • #35
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.
 
  • #36
Good. So what is your conclusion for part (b)?
 
  • #37
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.
 
  • #38
yes, I undestand, thank you very much...
 

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