Satellite Velocity Homework: Calculate Distance, Acceleration, and Speed

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SUMMARY

The discussion focuses on calculating the necessary parameters for a satellite orbiting Earth at a gravitational field strength of 4.5 N/kg. The calculated distance above Earth's surface for the satellite's orbit is 3.0 x 106 m. The correct orbital speed, after correcting for mathematical errors, is determined to be approximately 6509.68 m/s. The acceleration of the satellite is confirmed to be directed towards Earth, equivalent to 4.5 m/s2.

PREREQUISITES
  • Understanding of gravitational force equations, specifically Fg = G mE / r2
  • Knowledge of circular motion dynamics, including ac = v2/r
  • Familiarity with orbital mechanics and satellite motion
  • Proficiency in mathematical operations involving square roots and exponents
NEXT STEPS
  • Study the derivation of gravitational force equations in orbital mechanics
  • Learn about the relationship between gravitational acceleration and circular motion
  • Explore the concept of orbital speed and its calculation methods
  • Investigate the implications of altitude on satellite performance and stability
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Students in physics or engineering fields, particularly those focusing on orbital mechanics, satellite technology, and gravitational physics.

AClass
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Homework Statement



A satellite is designed to orbit Earth at an altitude above its surface that will place it in a gravitational field with a strength of 4.5 N/kg.

a) Calculate the distance above the surface of Earth at which the satellite must orbit.

b)Assuming the orbit is circular, calculate the acceleration of the satellite and its direction.

c)At what speed must the satellite travel in order to maintain this orbit?

mass of Earth : 5.98 x 10^24 kg
radius of the Earth : 6.38 x 10^6 m

Homework Equations



Fg= G mE / r^2
ac=v^2/r
v= sqrt ( GmE/r )
ac= 4Pi^2 r / T^2


The Attempt at a Solution



a)
4.5N/Kg=G(5.98x10^24kg)/r^2

r=9.4168x10^6m

h=re-r=3.0x10^6

Distance above the surface at which the satellite must orbit is 3.0x10^6m.

b)

No clue, I have searched many resources and they fail to give any clear ideas on how to tackle this question.

c)

Using

v=sqt(GMe/r)
v=sqt(G(5.98x10^24kg)/9.4168x10^6)
v=6.5097x10^6 m/s

Seems a bit too much. Could someone verify c) and I would appreciated any help with b), also verification of a) but I think I got it.
 
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AClass said:
v=sqt(GMe/r)
v=sqt(G(5.98x10^24kg)/9.4168x10^6)
v=6.5097x10^6 m/s

Seems a bit too much. Could someone verify c) and I would appreciated any help with b), also verification of a) but I think I got it.

Check your math for (c). You may have slipped up on the exponents.

For (b), consider N/kg by another name...
 
gneill said:
Check your math for (c). You may have slipped up on the exponents.

For (b), consider N/kg by another name...

c)
v=sqt(GMe/r)
v=sqt(G(5.98x10^24kg)/9.4168x10^6)
v=6509.6776m/s
I had forgotten to take the square root.

for b)

Other sources suggested that 4.5N/Kg is equal to 4.5m/s^2, that being its only acceleration. Seems that someone else who has taken my course before had confirmed that 4.5m/s^2 is not the correct acceleration.

I believe finding the acceleration from centreifual, and circular motions is my best bet.

However, I'm limited in terms of variables.
 
I suspect that 'other sources' failed to provide a direction for the acceleration.
 
gneill said:
I suspect that 'other sources' failed to provide a direction for the acceleration.

Yes, they failed to provide a direction. All that was explained is in the direction towards the earth.
 

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