E= 2.377074x1010 J * 380 kgE= 9.038314x1012 J

AI Thread Summary
The discussion focuses on calculating the total mechanical energy of a 380 kg satellite in a circular orbit at a distance of 3.0 Earth radii above the surface. Participants clarify that mechanical energy is the sum of kinetic and potential energy, and emphasize the need to use the correct radius for the orbit. The equations provided include gravitational potential energy and kinetic energy, with a discussion on how to derive the satellite's speed using the gravitational constant and the mass of the Earth. The final calculations lead to a specific mechanical energy value, which must be multiplied by the satellite's mass to find the total energy. The correct total mechanical energy is ultimately determined to be approximately 2.377074 x 10^10 J.
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Homework Statement


What is the total mechanical energy of a 380kg satellite in a circular orbit 3.0 Earth radii above the surface?


Homework Equations


W= E2- E1

E= 1/2 mv2- \frac{GmM}{r}


The Attempt at a Solution



I'm not sure if the equations above are suitable to solve this or I just don't understand how to start the question.
 
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Mechanical Energy is the sum of the potential energy, and the kinetic energy.

Emechanical = Ekinetic + EPotential

Since you're not close to the surface of the Earth, equating the potential energy to (mgh) is not applicable.

Epotential = (-GMm)/r

- r is the distance from the two objects centers,
- G is the gravitational constant of Earth
- M is the mass of Earth
- m is the mass of the satellite

Em = (1/2)mv2 - (GMm)/r

You have all the variables for the potential energy, and for the kinetic energy, you have the mass. So you need to solve for the velocity.

Since the mass of the satellite is just about negligible in relation to the earth, you can use the equation,

v = √(GM/r)

With that, you should be able to solve for the total mechanical energy.
 
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hsphysics2 said:

Homework Statement


What is the total mechanical energy of a 380kg satellite in a circular orbit 3.0 Earth radii above the surface?

Homework Equations


[STRIKE]W= E2- E1[/STRIKE]

E= 1/2 mv2- \frac{GmM}{r}

The Attempt at a Solution



I'm not sure if the equations above are suitable to solve this or I just don't understand how to start the question.

The equation is good, you need to find the speed of the satellite. The satellite travels along a circle with speed v. What is the radius of the orbit? What is its centripetal acceleration? What force provides the centripetal force?

ehild
 
ehild said:
The equation is good, you need to find the speed of the satellite.


The satellite travels along a circle with speed v. What is the radius of the orbit? What is its centripetal acceleration? What force provides the centripetal force?

ehild

so,

Fc= mac
(GmM)/RE2=(mv2)/RE
v2=(GM)/RE

where RE is the radius of the earth


then,

E= K+ UG where K=1/2(mv2) and UG=-(GmM)/r2 and r= 2.55x107m (equal to 4 Earth radii)

so,

E= 1/2(mv2) - (GmM)/r2
E= 1/2(GM)/RE) - (GM)/r2
E= 15633739.23 J
 
hsphysics2 said:
so,

Fc= mac
(GmM)/RE2=(mv2)/RE
v2=(GM)/RE

where RE is the radius of the earth

Why do you calculate the speed of the satellite with the radius of Earth?? You know that the radius of the the circular orbit is 4 Earth-radius.

ehild
 
ehild said:
Why do you calculate the speed of the satellite with the radius of Earth?? You know that the radius of the the circular orbit is 4 Earth-radius.

ehild

so it would just be

E= 1/2(GM/r) - (GM)/r
E= -7819322.353 J
 
hsphysics2 said:
so it would just be

E= 1/2(GM/r) - (GM)/r
E= -7819322.353 J

You haven't included the mass of the satellite. So as it stands so far, what you've calculated is the Specific Mechanical Energy (energy per kg).
 
gneill said:
You haven't included the mass of the satellite. So as it stands so far, what you've calculated is the Specific Mechanical Energy (energy per kg).

would it be,

E= K + UG
E= 1/2 (mv2) - (GmM)/r
E= 1/2 m(GM/r) - (GmM)/r
E= 2.377074x1010 J
 
hsphysics2 said:
would it be,

E= K + UG
E= 1/2 (mv2) - (GmM)/r
E= 1/2 m(GM/r) - (GmM)/r
E= 2.377074x1010 J

No. The formula's okay, but something went wrong in the execution.

Your previous value for the specific energy was good. Just multiply that by the mass of the satellite!
 
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