Calculate Satellite Velocity at 3600 km Above Earth

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SUMMARY

The discussion focuses on calculating the velocity of a satellite orbiting 3600 km above Earth's surface. The correct formula to use is v = √(GM/r), where G is the gravitational constant, M is the mass of Earth, and r is the distance from the center of Earth. Participants noted that the radius should be adjusted to account for Earth's radius, leading to a value greater than the initially calculated 5940 m/s. It is emphasized that the acceleration due to gravity decreases with altitude, affecting the final velocity calculation.

PREREQUISITES
  • Understanding of gravitational force and centripetal acceleration
  • Familiarity with the formula v = √(GM/r)
  • Knowledge of Earth's radius and gravitational constant values
  • Basic algebra for manipulating equations
NEXT STEPS
  • Research the gravitational constant (G) and its significance in orbital mechanics
  • Learn about the effects of altitude on gravitational acceleration
  • Study the concept of centripetal acceleration in circular motion
  • Explore satellite motion and orbital dynamics in physics
USEFUL FOR

Students in physics or engineering, educators teaching orbital mechanics, and anyone interested in satellite dynamics and gravitational calculations.

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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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