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

In summary, the total mechanical energy of a 380kg satellite in a circular orbit 3.0 Earth radii above the surface is approximately 5.94736x1013 J. This is found by calculating the specific mechanical energy of the satellite, which is -7819322.353 J/kg, and then multiplying it by the mass of the satellite.
  • #1
hsphysics2
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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- [itex]\frac{GmM}{r}[/itex]


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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  • #2
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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  • #3
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- [itex]\frac{GmM}{r}[/itex]

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
 
  • #4
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
 
  • #5
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
 
  • #6
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
 
  • #7
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).
 
  • #8
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
 
  • #9
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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FAQ: E= 2.377074x1010 J * 380 kgE= 9.038314x1012 J

1. What is mechanical energy in relation to orbit?

Mechanical energy refers to the sum of potential energy and kinetic energy in a system. In the context of orbit, mechanical energy is the combination of the gravitational potential energy and the kinetic energy of a celestial body as it moves along its orbital path.

2. How is mechanical energy conserved in orbit?

The law of conservation of energy states that energy cannot be created or destroyed, but can only be transferred or converted from one form to another. In the case of orbit, mechanical energy is conserved because the gravitational potential energy is converted into kinetic energy as the body moves closer to the center of gravity, and vice versa.

3. How does mechanical energy affect the motion of a body in orbit?

Mechanical energy is directly related to the speed and distance of a body in orbit. As a body moves closer to the center of gravity, its potential energy decreases and its kinetic energy increases, causing it to speed up. As it moves further away, the opposite occurs and the body slows down.

4. Can mechanical energy be used to escape orbit?

Yes, mechanical energy can be used to escape orbit. In order for a body to leave its orbit, it must have enough kinetic energy to overcome the gravitational pull of the object it is orbiting. By increasing its mechanical energy, a body can achieve escape velocity and break free from orbit.

5. How is mechanical energy calculated in orbit?

The total mechanical energy of a body in orbit is calculated by adding together its potential energy and kinetic energy. The potential energy is calculated using the formula mgh, where m is the mass of the body, g is the gravitational acceleration, and h is the distance from the center of gravity. The kinetic energy is calculated using the formula 1/2mv^2, where m is the mass of the body and v is its velocity.

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