Minimum work required to put object into orbit?

In summary: You have only calculated the change in gravitational potential energy. How does the satellite stay in orbit? It has to have the right velocity to keep it in orbit, right? Actually, what you wrote is pretty close to what I was looking for, depending no how I interpret your notation...F(mg)=mv^2/rBut remember that "g" is the acceleration of gravity at the surface of the Earth. What is the more general equation for the gravitational force farther away from the Earth?
  • #1
Bekamop99
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Homework Statement


Scientists are planning to launch a rocket from the surface of the Earth into an orbit at a distance of 18000 km above the centre of the Earth. The radius of the Earth is 6400 km and it has mass 6.0x10^24 kg.
What is the minimum work done to move the 150kg mass of the rocket into this orbit?

Homework Equations

The Attempt at a Solution


distance = 18x10^6 + 6.4x10^6 = 2.44x10^7
mass Earth = 6x10^24
mass satellite = 150kg
energy = (-GMm)/r = 2.46 x10^9

My answer is very wrong! Please help! Thank you :)
 
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  • #2
Bekamop99 said:

Homework Statement


Scientists are planning to launch a rocket from the surface of the Earth into an orbit at a distance of 18000 km above the centre of the Earth. The radius of the Earth is 6400 km and it has mass 6.0x10^24 kg.
What is the minimum work done to move the 150kg mass of the rocket into this orbit?

Homework Equations

The Attempt at a Solution


distance = 18x10^6 + 6.4x10^6 = 2.44x10^7
mass Earth = 6x10^24
mass satellite = 150kg
energy = (-GMm)/r = 2.46 x10^9

My answer is very wrong! Please help! Thank you :)
Welcome to the PF.:smile:

You have only calculated the change in gravitational potential energy. How does the satellite stay in orbit? It has to have the right velocity to keep it in orbit, right? Show the Relevant Equations and your work on the orbital velocity for that height, and then use that to add in the KE component of the satellite's energy...

EDIT -- I'm not actually sure what you did calculate...
 
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  • #3
Bekamop99 said:
energy = (-GMm)/r = 2.46 x10^9
That formula gives the gravitational potential energy for a given separation distance. (Note that it should be negative, since it assumes a reference point of infinite separation.) You need to compare the change in total energy of the satellite as it is moved from the Earth surface to the orbiting radius.
 
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  • #4
Doc Al said:
That formula gives the gravitational potential energy for a given separation distance. (Note that it should be negative, since it assumes a reference point of infinite separation.) You need to compare the change in total energy of the satellite as it is moved from the Earth surface to the orbiting radius.

Thanks for this but I can't work out how to find Kinetic energy. I tried F(mg)=mv^2/r , finding v at max and min orbits.
Then I added (Ep max orbit + Ek max orbit) - (Ep min orbit + Ek min orbit). But that's not right! I think I've got this really confused. I may just ask my teacher in a few days time! :)
 
  • #5
Bekamop99 said:
Thanks for this but I can't work out how to find Kinetic energy. I tried F(mg)=mv^2/r , finding v at max and min orbits.
Then I added (Ep max orbit + Ek max orbit) - (Ep min orbit + Ek min orbit). But that's not right! I think I've got this really confused. I may just ask my teacher in a few days time! :)
So what did you get for the change in gravitational potential energy (GPE) between the Earth's surface and the orbit height? Please show your work.

And what equation relates the force on an object in orbit to its distance from the center of the Earth? And what equation relates the radius of centripital motion to the velocity of the object in motion and the centripital force? Use those equations to find the velocity v. Please show your work. :smile:
 
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  • #6
berkeman said:
And what equation relates the force on an object in orbit to its distance from the center of the Earth?
Actually, what you wrote is pretty close to what I was looking for, depending no how I interpret your notation...
Bekamop99 said:
F(mg)=mv^2/r
But remember that "g" is the acceleration of gravity at the surface of the Earth. What is the more general equation for the gravitational force farther away from the Earth?
 
  • #7
Bekamop99 said:
Thanks for this but I can't work out how to find Kinetic energy. I tried F(mg)=mv^2/r , finding v at max and min orbits.
Forget about max & min orbits (unless your teacher discusses such). Just assume a simple circular orbit at the given distance. That equation, assuming you use the correct value for gravitational force (not mg!) will give you the kinetic energy.
 
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  • #8
berkeman said:
So what did you get for the change in gravitational potential energy (GPE) between the Earth's surface and the orbit height? Please show your work.

And what equation relates the force on an object in orbit to its distance from the center of the Earth? And what equation relates the radius of centripital motion to the velocity of the object in motion and the centripital force? Use those equations to find the velocity v. Please show your work. :smile:

g=
berkeman said:
So what did you get for the change in gravitational potential energy (GPE) between the Earth's surface and the orbit height? Please show your work.

And what equation relates the force on an object in orbit to its distance from the center of the Earth? And what equation relates the radius of centripital motion to the velocity of the object in motion and the centripital force? Use those equations to find the velocity v. Please show your work. :smile:

Thank you both for your responses. I've realized I read a few things wrong in the question in the first place so have had another look at it but am still 4 decimal places out! Attached is my working, see what you can make of it! But I really understand if you would rather not reply! :)
15934571_724955874324728_1166831636_n.jpg
 
  • #9
Can you clear up your image and repost it? It's hard to read the way it is. Better yet would be to type your work into the forum window... :smile:
 
  • #10
Here's a brightened copy, but you should still get used to typing your work into the forum. It makes it much easier for us to read and comment on.

15934571_724955874324728_1166831636_n.jpg
 
  • #11
Looks to me like you're computing ΔPE. What about KE? (I think you can safely ignore the Earth's rotation.)
 

Related to Minimum work required to put object into orbit?

1. What is the definition of "minimum work required to put an object into orbit?"

The minimum work required to put an object into orbit refers to the amount of energy needed to overcome Earth's gravitational pull and achieve a stable orbit around the planet.

2. How is the minimum work calculated for putting an object into orbit?

The minimum work required to put an object into orbit is calculated using the formula W = GMm(1/r - 1/2a), where W is the work done, G is the gravitational constant, M and m are the masses of the two objects, r is the distance between them, and a is the radius of the larger object.

3. What factors affect the minimum work required to put an object into orbit?

The minimum work required to put an object into orbit is affected by the mass and radius of the planet, the mass and velocity of the object, and the distance between the planet and the object.

4. Can the minimum work required to put an object into orbit be reduced?

Yes, the minimum work required to put an object into orbit can be reduced by increasing the initial velocity of the object or by using a more efficient propulsion system.

5. How does the minimum work required to put an object into orbit vary for different planets?

The minimum work required to put an object into orbit varies for different planets depending on their mass and radius. The larger and more massive a planet is, the more work is required to put an object into orbit around it.

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