How High Must a Satellite Orbit to Maintain Specific Gravitational Strength?

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SUMMARY

A satellite must orbit Earth at an altitude of 3.02 x 106 m to experience a gravitational strength of 4.5 N/kg. The calculations utilize the gravitational force equation Fg = G mE / r2 to determine the required distance from the Earth's center. The satellite must travel at a speed of 11,489.16 m/s to maintain this orbit, calculated using v = sqrt(GmE / r). For determining the acceleration and direction, Newton's second law is recommended as a method to solve part b of the problem.

PREREQUISITES
  • Understanding of gravitational force equations, specifically Fg = G mE / r2
  • Knowledge of centripetal acceleration formulas, such as ac = v2 / r
  • Familiarity with Newton's second law of motion
  • Basic proficiency in algebra and square root calculations
NEXT STEPS
  • Study the application of Newton's second law in circular motion scenarios
  • Learn about gravitational fields and their effects on satellite motion
  • Explore the relationship between orbital speed and gravitational force
  • Investigate the implications of varying altitudes on satellite stability and speed
USEFUL FOR

Students studying physics, particularly those focusing on gravitational forces and satellite dynamics, as well as educators seeking to enhance their understanding of orbital mechanics.

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Homework Statement


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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?

Given values in the book:
mass of Earth : 5.98 x 10^24 kg
radius of the Earth : 6.38 x 10^6 m
g : 6.67 x 10^-11 Nm^2/kg^2

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) So we have to find the distance of the satellite above the surface of the Earth and for that we will find its distance from the centre of the Earth and then subtract it from the Earth's radius to obtain the final result.

Fg= GmE/r^2
4.5=(6.67x10^-11)(5.98x10^24) / r^2
r^2= 3.98866x10^14 / 4.5
r= sqrt (8.8637x10^13)
r=9.4x10^6 m

r surface = r total - r earth
r surface = 9.4x10^6 - 6.38x10^6
r surface = 3.02x10^6 m

The satellite must orbit at 3.02x10^6 m from the surface of the earth.

b)
This is the one i don't get it althought I'm not sure which one the centripetal acceleration formulas to use, i have no idea how to figure out its direction, please help !

c)

v= sqrt (GmE / r)
v=sqrt ( (6.67x10^-11)(5.98x10^24)/3.02x10^6)
v=sqrt (3.98866x10^14/3.02x10^6)
v=sqrt (1.32x10^8)
v=11489.16 m/s

The satellite must travel at 11489.16 m/s in order to maintain its orbit.


again if possible to help me with question b) and perhaps review my work for questions a) and c) , thank you everyone!
 
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The principles of a and c are right, so as long as you did the numbers okay, those looks good.

For part b), try using Newton's second law since you don't know the period. :)
 

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