Satellite Orbit Around Mars: Force & Speed Calculation

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SUMMARY

The discussion focuses on calculating the gravitational force and orbital speed of a 2500 kg satellite intended to orbit Mars at a distance of 1.8 times the radius of Mars. The gravitational force calculation requires adding the radius of Mars to the distance above the surface, leading to a corrected formula. The gravitational force was initially miscalculated as 2864.72 N, but the correct approach involves using the total distance from the center of Mars. For orbital speed, the centripetal acceleration formula is applicable, emphasizing the relationship between speed and radius in uniform circular motion.

PREREQUISITES
  • Understanding of Newton's Law of Universal Gravitation
  • Familiarity with centripetal acceleration concepts
  • Knowledge of gravitational constant (G = 6.67428 x 10^-11 N-m²/kg²)
  • Basic algebra for manipulating equations
NEXT STEPS
  • Calculate gravitational force using the correct distance formula for satellite orbit
  • Learn about centripetal acceleration and its application in orbital mechanics
  • Explore the implications of varying orbital radii on satellite speed
  • Study the effects of gravitational forces on satellite stability and trajectory
USEFUL FOR

Astronomy enthusiasts, physics students, aerospace engineers, and anyone interested in satellite dynamics and orbital mechanics will benefit from this discussion.

mrshappy0
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Scientists want to place a 2500 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 1.8 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:

mmars = 6.4191 x 1023 kg
rmars = 3.397 x 106 m
G = 6.67428 x 10-11 N-m2/kg2

1)What is the force of attraction between Mars and the satellite?
I got 2864.722760095893N but the online HW program says I am wrong and that I need to add one more radius. This is the formula I used: (2500(6.674282(10^-11))(6.4191(10^23)))/(1.8(3.397(10^6)))^2

2)What speed should the satellite have to be in a perfectly circular orbit?

No idea where to start with this one. Should I start with a free body diagram. I think I would have the W vector pointing in towards the center of Mars and I would have to make the movement along the vertical axis (y) 0 and the movement along the horizontal axis (x) constant. Correct?
 
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This isn't a HW problem. I am just studying. I have spent a fair amount of time on this and can't get it right.
 
mrshappy0 said:
Scientists want to place a 2500 kg satellite in orbit around Mars. They plan to have the satellite orbit a distance equal to 1.8 times the radius of Mars above the surface of the planet. Here is some information that will help solve this problem:

mmars = 6.4191 x 1023 kg
rmars = 3.397 x 106 m
G = 6.67428 x 10-11 N-m2/kg2

1)What is the force of attraction between Mars and the satellite?
I got 2864.722760095893N but the online HW program says I am wrong and that I need to add one more radius. This is the formula I used: (2500(6.674282(10^-11))(6.4191(10^23)))/(1.8(3.397(10^6)))^2

2)What speed should the satellite have to be in a perfectly circular orbit?

No idea where to start with this one. Should I start with a free body diagram. I think I would have the W vector pointing in towards the center of Mars and I would have to make the movement along the vertical axis (y) 0 and the movement along the horizontal axis (x) constant. Correct?

Welcome to the PF.

You are correct to post here in the HH forums -- all schoolwork-type problems should be posted here. And it is good that you are showing your work.

The questions uses the words "above the surface of the planet", so that's why you need to add in one radius of the planet to get your attractive force calculation based on the distance between the centers of mass of the two objects.

And for the orbital speed, are you familiar with the equation(s) for uniform circular motion due to an attractive force?
 
Q1: I plugged in a 2 for another radius and got this: (2500(6.674282(10^-11))(6.4191(10^23)))/(1.8(2(3.397(10^6))))^2 = 716.1806900239733. The answer seems to be incorrect again. puzzled. I must be reading something wrong.

Q2: Uniform circular motion- so use centripetal acceleration (a=(v^2)/Radius)?
 
mrshappy0 said:
Q1: I plugged in a 2 for another radius and got this: (2500(6.674282(10^-11))(6.4191(10^23)))/(1.8(2(3.397(10^6))))^2 = 716.1806900239733. The answer seems to be incorrect again. puzzled. I must be reading something wrong.

Q2: Uniform circular motion- so use centripetal acceleration (a=(v^2)/Radius)?

Q1: The satellite will be 1.8 times the radius of Mars *above* Mars. So the total distance is not 2*1.8 rmars... What is it?

(And there is no reason to show more than 3-4 digits past the decimal point in your answers.)

Q2: Correct! Keep on going!
 
Q1: simple. Thanks.

Q2: Wow physics just baffles me. It's like magic. Lately it is just clicking for me and it never has until today. Knock on wood-I know it gets harder but it is a start. Thanks for your time.
 
mrshappy0 said:
Q1: simple. Thanks.

Q2: Wow physics just baffles me. It's like magic. Lately it is just clicking for me and it never has until today. Knock on wood-I know it gets harder but it is a start. Thanks for your time.

Glad that it's starting to click, and glad to help. :smile:
 

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