Satellite Orbit Around Mars: Force & Speed Calculation

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Homework Help Overview

The discussion revolves around calculating the gravitational force and orbital speed of a satellite intended to orbit Mars. The satellite's mass is given as 2500 kg, and it is to orbit at a distance of 1.8 times the radius of Mars above the planet's surface. Relevant constants such as Mars' mass and radius, as well as the gravitational constant, are provided.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of gravitational force, questioning the need to include the radius of Mars in the distance calculation. There are attempts to clarify the total distance for the force calculation and the correct application of formulas for circular motion.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning the correctness of their results. Some guidance has been provided regarding the interpretation of the problem's wording and the application of relevant physics concepts, particularly in relation to gravitational force and orbital speed.

Contextual Notes

There is a noted confusion regarding the interpretation of the distance for the gravitational force calculation, specifically how to account for the radius of Mars in relation to the satellite's orbit. Participants are also navigating the complexities of uniform circular motion and centripetal acceleration.

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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