Calculate Satellite Velocity at 3600 km Above Earth

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Homework Help Overview

The discussion revolves around calculating the velocity of a satellite positioned 3600 km above the Earth's surface, focusing on the implications of gravitational acceleration at that altitude and the appropriate radius for calculations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the original poster's calculation attempt and question the accuracy of the values used, particularly regarding gravitational acceleration and the radius in the context of circular motion.

Discussion Status

Some participants express agreement with the original poster's approach while others suggest that additional considerations regarding gravitational changes at altitude and the correct radius need to be addressed. Multiple interpretations of the problem are being explored.

Contextual Notes

There is an emphasis on the need to account for variations in gravitational acceleration at higher altitudes and the correct radius for the satellite's orbit, which may not simply be the altitude above the Earth's surface.

pb23me
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1. Homework Statement [/b]
calculate the velocity of a satellite 3600 km above the surface of the earth.


Homework Equations


Fnet= mg=ma
mg=ma
g=V2/r
r=3600000


The Attempt at a Solution

5940 m/s I am pretty sure this is correct just want to make sure.
 
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Looks good to me as long as you plugged in all the right values
 
Two things you need to include in your calculation:
1) at 3600 km above the surface of Earth, the acceleration due to gravity is not same as at the surface.
2) The radius in the circular motion equation is not 3600 km.
The correct value is a bit more than what you got.
 
Filip Larsen said:
Two things you need to include in your calculation:
1) at 3600 km above the surface of Earth, the acceleration due to gravity is not same as at the surface.
2) The radius in the circular motion equation is not 3600 km.
The correct value is a bit more than what you got.

You are right...but the equations are right, and as we all know that is all that matters once you start grad school :P haha
 
Just equate centripetal acceleration to the gravitational force at that height.
Better just remember the formula instead of deriving each time -

v = sq root (GM/r)

where G = grav constant, M= mass of earth, r = distance from centre of earth
 

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