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Lexington
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FYI: I'm new here but I'm just about done phys 20, just nearing the end of the course. We're working on calculating Work right now. More specifically calculating Mechanical Energy, Kinetic Energy and Potential Energy. I think I've pretty much got most of the equation solved, it's just one last detail I can't seem to figure out on my own. Now for the problem:
*NONE (ZERO) FRICTION (AKA ISOLATED) SYSTEM*
5. To maintain a circular orbit at a height of 180 km above Earth's surface, a satellite with a mass of 1.0 x 10^3 kg must travel at a constant speed of 7.8 km/s. Using an average value for g of 9.5 m/s^2, calculate the gravitational potential energy of the satellite, its kinetic energy, its mechanical energy, and the height it can rise to before losing all its kinetic energy.
Relevant equations
Ek = 1/2mv^2
Ep = mgh + zero
Em = Ek + Ep
Where:
Ek = Kinetic Energy
Ep = Potential Energy
Em = Mechanical Energy
m = Mass
v = Velocity
g = Gravity
h = Height (distance from Earth's surface)
Here is what I have so far. *IF THERE ARE ANY ERRORS PLEASE LET ME KNOW* Look below to see what I need help with.
h = 180 000m
m = 1.0x10^3kg
v = 7.8x10^3m/s
g = 9.5m/s^2
Ek = 1/2mv^2
= 1/2(1.0*10^3kg)(7.8x10^3m/s)^2
= 30,420,000,000
Ek = 3.0x10^10 J
The Kinetic Energy for the satellite is 3.0x10^10 J.Ep = mgh + zero
= (1.0x10^3kg)(9.5m/s^2)(180000m) + 0
= 1,710,000,000
= 1.7x10^9 J
The Gravitational Potential Energy of the satellite is 1.7x10^9 J.
Em = Ek + Ep
= (3.0x10^10 J) + (1.7x10^9 J)
= 301,700,000,000
= 3.0x10^11 J
The total Mechanical Energy for the satellite is 3.0x10^11 J.
The last part of the question asks for "the height it can rise to before losing all its kinetic energy." Now, I know that when the kinetic energy is 0 the potential energy is equal to the mechanical energy. So this is the incomplete equation I figured:
When Ek = 0
Em = 3.0x10^11 J
m = 1.0x10^3kg
g = ??
Em = Ep
= mgh
h = (Em)/(mg)
= (3.0x10^11J)/[(1.0x10^3kg)(____)
When we go farther away from earth, gravity Decreases. So evidently my input for gravity should be something else, something dependant on the height. So if not this equation, what equation should I use? Or IF this equation, what variable for gravity do I use, 9.5m/s^2?
*NONE (ZERO) FRICTION (AKA ISOLATED) SYSTEM*
5. To maintain a circular orbit at a height of 180 km above Earth's surface, a satellite with a mass of 1.0 x 10^3 kg must travel at a constant speed of 7.8 km/s. Using an average value for g of 9.5 m/s^2, calculate the gravitational potential energy of the satellite, its kinetic energy, its mechanical energy, and the height it can rise to before losing all its kinetic energy.
Relevant equations
Ek = 1/2mv^2
Ep = mgh + zero
Em = Ek + Ep
Where:
Ek = Kinetic Energy
Ep = Potential Energy
Em = Mechanical Energy
m = Mass
v = Velocity
g = Gravity
h = Height (distance from Earth's surface)
Here is what I have so far. *IF THERE ARE ANY ERRORS PLEASE LET ME KNOW* Look below to see what I need help with.
h = 180 000m
m = 1.0x10^3kg
v = 7.8x10^3m/s
g = 9.5m/s^2
Ek = 1/2mv^2
= 1/2(1.0*10^3kg)(7.8x10^3m/s)^2
= 30,420,000,000
Ek = 3.0x10^10 J
The Kinetic Energy for the satellite is 3.0x10^10 J.Ep = mgh + zero
= (1.0x10^3kg)(9.5m/s^2)(180000m) + 0
= 1,710,000,000
= 1.7x10^9 J
The Gravitational Potential Energy of the satellite is 1.7x10^9 J.
Em = Ek + Ep
= (3.0x10^10 J) + (1.7x10^9 J)
= 301,700,000,000
= 3.0x10^11 J
The total Mechanical Energy for the satellite is 3.0x10^11 J.
The last part of the question asks for "the height it can rise to before losing all its kinetic energy." Now, I know that when the kinetic energy is 0 the potential energy is equal to the mechanical energy. So this is the incomplete equation I figured:
When Ek = 0
Em = 3.0x10^11 J
m = 1.0x10^3kg
g = ??
Em = Ep
= mgh
h = (Em)/(mg)
= (3.0x10^11J)/[(1.0x10^3kg)(____)
When we go farther away from earth, gravity Decreases. So evidently my input for gravity should be something else, something dependant on the height. So if not this equation, what equation should I use? Or IF this equation, what variable for gravity do I use, 9.5m/s^2?
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