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

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Homework Help Overview

The discussion revolves around calculating the energy required for a satellite to reach a circular orbit at an altitude of 300 km above the Earth's surface. The problem also includes considerations of the satellite's velocity changes over time and the effects of atmospheric resistance as it approaches the Earth.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the initial kinetic energy of the satellite at launch and whether it should be included in energy calculations. There are questions about the significance of the Earth's rotation and the change in kinetic energy due to the satellite's mass relative to the Earth.
  • Some participants explore the relationship between radial and tangential components of velocity, particularly in the context of the satellite's gradual descent and the comparison of these velocities over time.
  • There are attempts to calculate the average resistance force acting on the satellite and to verify assumptions regarding the satellite's motion and energy changes.

Discussion Status

The discussion is active, with participants providing insights and calculations related to the energy and velocity components of the satellite. There is a focus on verifying assumptions and clarifying concepts, particularly regarding the comparison of radial and tangential velocities. Guidance has been offered on unit conversions and the significance of different energy components.

Contextual Notes

Participants are working within the constraints of a homework assignment, which may limit the information available for calculations. The discussion includes a focus on verifying calculations and assumptions, particularly regarding the satellite's motion and the effects of atmospheric resistance.

  • #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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