HELP! ~ Mechanical Energy vs Potential Energy & Kinetic Energyby Lexington Tags: energy, kinetic, mechanical, potential 

#1
May1408, 04:21 AM

P: 4

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? 



#2
May1408, 04:44 AM

Emeritus
Sci Advisor
PF Gold
P: 9,789

Hi Lexington and welcome to PF,
Firstly, I want to thank you for submitting such a detailed question and solution. Secondly, although I haven't checked your arithmetic your methods seem correct. Thirdly, the question states that you may use an average value for the acceleration due to gravity. Therefore, I would think it was safe to assume that it would be valid to use the averaged value of g in the final part of the question. Your formula is correct. 



#3
May1408, 05:03 AM

P: 4

You're welcome =)
And thank you for all that you said. I really appreciate the help. 



#4
Jul2708, 08:53 PM

P: 33

HELP! ~ Mechanical Energy vs Potential Energy & Kinetic Energy
You should use the gravitational formula Fg=G(m1m2)/r2 where G is the gravitational constant of 6.67*10^11, Fg is the force of Gravity so divide by the mass to get acceleration, mass1 is the mass of the earth, mass two is the mass of the satelite, and r^2 is the radius from the center of the earth squared so you take the earths radius and add the height of the satellite then square the sum. That should give you the closest approximation, you can also use that formula to determine escape velocity. Keep in mind the error due to air resistance and terminal velocity, but that is not important until calculus.




#5
Jul2708, 09:00 PM

P: 33

never mind what I said the 9.5m/s^2 is what they expect you to use.



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