Planets and satellites, law of periods circular orbit.

In summary: However, Fg /m gives the acceleration of the satellite due to the attraction of the planet. This is what you would call the "acceleration due to gravity of the planet". You are given the acceleration due to the force of gravity (Fg) of the satellite, which is the same as the acceleration due to the mass (M) of the satellite.
  • #1
J-dizzal
394
6

Homework Statement


A 15 kg satellite has a circular orbit with a period of 4.4 h and a radius of 3.5 × 106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 1.2 m/s2, what is the radius of the planet?

Homework Equations

The Attempt at a Solution


20150719_221818_zpsyk9ccfhe.jpg

I think I am on the right track, but I am stuck finding the radius of the planet. in my equation at the bottom for radius what would i use for M and m?
 
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  • #2
In your equation ##r=\sqrt{\frac{GMm}{F_g}}## r is the distance between the two objects, not the radius of the planet.

Try to use the centripetal acceleration to find the mass of the planet (only 1 mass will cause this orbit) and then try to find the radius (with the mass known, only 1 radius will cause that value of surface gravity).
 
Last edited:
  • #3
Considering Newton's 2nd law, what does Fg /m represent?
 
  • #4
Nathanael said:
In your equation r=GMmFg−−−−√r=\sqrt{\frac{GMm}{F_g}} r is the distance between the two objects, not the radius of the planet. That equation is really of no use.
would it work if M is the mass of the planet and m is a mass on the surface of the planet? the distance between them would be about the radius of the planet?

TSny said:
Considering Newton's 2nd law, what does Fg /m represent?
acceleration. But in my equation for r; Fg is the force of acceleration on the surface of the planet. would that work with the given a=1.2m/s2
 
  • #5
J-dizzal said:
would it work if M is the mass of the planet and m is a mass on the surface of the planet? the distance between them would be about the radius of the planet?
Oh my bad, yes then it would work. You just need to find the mass of the planet first.
 
  • #6
Nathanael said:
Oh my bad, yes then it would work. You just need to find the mass of the planet first.
Does my equation for M look ok, toward the top of my page M=946496.4kg?
If its ok, then would any value for m work in my equation for r?
 
  • #7
J-dizzal said:
Does my equation for M look ok, toward the top of my page M=946496.4kg?
If its ok, then would any value for m work in my equation for r?
Sorry for ignoring your initial work, apparently I'm not so good at skimming pictures o0)

The equation is right but you didn't cube the radius when you solved it.
 
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  • #8
Nathanael said:
Sorry for ignoring your initial work, apparently I'm not so good at skimming pictures o0)

The equation is right but you didn't cube the radius when you solved it.
In my equation for r, I am letting Fg=a. where a=1.2m/s/s is a given in the problem statement. I am guessing this is wrong because its the force on the satellite at distance r from the surface.
 
  • #9
J-dizzal said:
In my equation for r, I am letting Fg=a. where a=1.2m/s/s is a given in the problem statement. I am guessing this is wrong because its the force on the satellite at distance r from the surface.
a is 1.2 m/s/s, not Newtons. Consider an object on the surface of mass m. If it accelerates at 1.2m/s/s, then what must the force on that object (of mass m) be?
 
  • #10
J-dizzal said:
acceleration. Yes, But in my equation for r; Fg is the force of acceleration on the surface of the planet. would that work with the given a=1.2m/s2
Fg = GMm/r2 gives the force of gravity on m as a function of r. Using this, you can derive an expression for FG /m as a function of r. Thus, you have an expression for the acceleration due to gravity as a function of r. Apply this for a point at the surface of the planet.
 
  • #11
Nathanael said:
a is 1.2 m/s/s, not Newtons. Consider an object on the surface of mass m. If it accelerates at 1.2m/s/s, then what must the force on that object (of mass m) be?
F=(15kg)(1.2m/s/s)=18N

TSny said:
Fg = GMm/r2 gives the force of gravity on m as a function of r. Using this, you can derive an expression for FG /m as a function of r. Thus, you have an expression for the acceleration due to gravity as a function of r. Apply this for a point at the surface of the planet.
Fg/M = Gm/r2 so this would be the acceleration of M?
 
  • #12
J-dizzal said:
Fg/M = Gm/r2 so this would be the acceleration of M?

Well, yes that would give the acceleration of the planet (M) due to the attraction of the satellite (m). But that would be extremely small and it is not of relevance to this problem.

However, Fg /m gives the acceleration of the satellite due to the attraction of the planet. This is what you would call the "acceleration due to gravity of the planet". You are given the acceleration due to gravity of the planet at the surface of the planet.

So you want to find an expression for Fg /m rather than Fg /M. Then apply the expression to a point near the surface of the planet.
 
  • #13
TSny said:
Well, yes that would give the acceleration of the planet (M) due to the attraction of the satellite (m). But that would be extremely small and it is not of relevance to this problem.

However, Fg /m gives the acceleration of the satellite due to the attraction of the planet. This is what you would call the "acceleration due to gravity of the planet". You are given the acceleration due to gravity of the planet at the surface of the planet.

So you want to find an expression for Fg /m rather than Fg /M. Then apply the expression to a point near the surface of the planet.
im getting R= sqrt(GM/a) =1.337x106 same answer i got before...
 
  • #14
R = sqrt(GM/a) is correct. What did you get for the mass of the planet, M, after correcting for cubing r as pointed out by Nathanael in post #7?
 
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  • #15
TSny said:
R = sqrt(GM/a) is correct. What did you get for the mass of the planet, M, after correcting for cubing r as pointed out by Nathanael in post #7?
I got 3.219x1022kg
 
  • #16
TSny said:
R = sqrt(GM/a) is correct. What did you get for the mass of the planet, M, after correcting for cubing r as pointed out by Nathanael in post #7?
ok i got it correct now thanks again.
 

What is the law of periods for a circular orbit?

The law of periods states that the square of a planet's orbital period is directly proportional to the cube of its average distance from the sun. This means that the farther a planet is from the sun, the longer its orbital period will be.

How does the law of periods apply to satellites?

The law of periods also applies to satellites orbiting a planet. In this case, the square of the satellite's orbital period is directly proportional to the cube of its average distance from the planet. This means that the farther a satellite is from the planet, the longer its orbital period will be.

Why is the law of periods important in astronomy?

The law of periods allows scientists to predict the orbital period of a planet or satellite based on its distance from the sun or planet. This is crucial in understanding the dynamics of our solar system and in studying other planetary systems in the universe.

Does the law of periods apply to all types of orbits?

The law of periods only applies to circular orbits. For elliptical orbits, the law of periods is modified to include the eccentricity of the orbit. This means that the orbital period of a planet or satellite in an elliptical orbit will vary depending on its distance from the sun or planet at different points in its orbit.

How was the law of periods discovered?

The law of periods was first discovered by German astronomer Johannes Kepler in the early 17th century. He observed that the orbital period of a planet was related to its distance from the sun, and through further calculations and observations, he developed the law of periods. Later, Isaac Newton's theory of gravitation provided a mathematical explanation for the law of periods.

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