Time it takes for a satellite to orbit the Earth

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SUMMARY

The time it takes for a satellite to orbit the Earth in a near-Earth orbit can be calculated using gravitational equations. The gravitational acceleration is defined by the formula g = GM/r², where M is the mass of the Earth (5.98x10²⁴ kg) and r is the radius of the Earth (6.38x10⁶ m). The velocity of the satellite is determined by setting the centripetal acceleration equal to gravitational acceleration, leading to a velocity of approximately 1.99x10⁷ m/s. The corrected gravitational acceleration is approximately 9.796 m/s², confirming that the mass of the satellite does not affect the orbital period.

PREREQUISITES
  • Understanding of Newton's law of universal gravitation
  • Familiarity with centripetal acceleration concepts
  • Knowledge of gravitational acceleration calculations
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of orbital mechanics equations
  • Learn about Kepler's laws of planetary motion
  • Explore the effects of altitude on satellite velocity and period
  • Investigate the differences between circular and elliptical orbits
USEFUL FOR

Students studying physics, aerospace engineers, and anyone interested in satellite dynamics and orbital mechanics.

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Homework Statement



Determine the time it takes for a satellite to orbit the Earth in a circular "near-Earth" orbit. The definition of "near-Earth" orbit is one which is at a height above the surface of the Earth which is small compared to the radius of the Earth, so that you may take the acceleration due to gravity as essentially the same as that on the surface. Does your result depend on the mass of the satellite?

M(earth)= 5.98x10^24 kg
R(earth)=6.38x10^6 m
G= 6.67x10^-11

Homework Equations



g= GM/r^2

FG= Gm1m2/r^2

The Attempt at a Solution



I am not sure where to start since I don't know what equation to use.
 
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One more equation might help you. The satellite in orbit will have an acceleration equal to a = \frac{v^2}{r} which will equal Earth's gravity, g = \frac{GM}{r^2}

Setting those equal to each other, can you determine the velocity of the satellite and use that to find the period?
 
The gravitational force between the Earth and the satellite provides the centripetal force of the satellite. Are you able to make an equation?
 
Okay, I'm not sure if this is right, but here's my attempt:

I did v^2/r = GM/r^2
& plugged in:
V^2/(6.38x10^6) = (6.67x10^-11)(5.98x10^24)/(6.38x10^6)

and ended up with:
V= 1.99x10^7 m/s

when i plugged it into a=v^2/r i got a= 6.25x10^7

this doesn't seem right, isn't is suppose to be near 9.8 m/s?
 
You didn't square the radius on the right side.
 
Wow, dumb mistake...I redid it by squaring it and got 9.796 m/s. Thank you for your help!
 

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