A 1 007-kg satellite orbits the Earth at a constant altitude of 94-km. How much
ID: 1268317 • Letter: A
Question
A 1 007-kg satellite orbits the Earth at a constant altitude of 94-km. How much energy must be added to the system to move the satellite into a circular orbit with altitude 208 km? How is the total energy of an object in circular orbit related to the potential energy? MJ What is the change in the system's kinetic energy? Is the satellite moving faster or slower when it's orbit is at a higher altitude? MJ What is the change in the system's potential energy? What is the equation for the gravitation a potential energy of s system of two objects? MJExplanation / Answer
A. There are no nonconservative forces acting on the satellite, so the mechanical energy is conserved going from orbit 1 into orbit 2 except for the added energy. Note that E1 = PE1 + KE1 and E2 = PE2 + KE2 where E is the mechanical energy, PE is the gravitational potential energy, and KE is the kinetic energy.
We have that PE = mg(R + h) where R is the radius of the Earth (= 6.38 x 10^6 m), h is the height of the orbit and m is the mass of the satellite. We can assume g does not change by going from orbit 1 to orbit 2 since the height difference is very small. Also we have KE = 1/2 m v^2 ignoring any rotational effects. It is given that h1 = 110 km and h2 = 210 km. But what are v1 and v2?
The velocity v can be calculated using Newton's Law of Universal Gravitation. Since F = ma by the 2nd law of dynamics, then F = G m M / (R + h)^2 = m v^2 / (R + h) = m a, where G is a universal constant, m is satellite's mass, M is the Earth's mass, and v^2 / (R + h) is the centripetal acceleration for the circular orbit.
If you do a little algebra, you can find that v = sqrt(GM/(R + h)). So to put it all together, we have E1 + added energy = E2 which gives mg(R + h1) + 1/2 m GM/(R + h1) + added energy = mg(R + h2) + 1/2 m GM/(R + h2). Put in the values and calculate the added energy.
added energy = mg(h2-h1)+1/2mGM(h1-h2/(R+h2)(R+h1))
added energy = 1007*9.8*(208-94)+(0.5*1007*6.67*10^-11*5.98 x 10^24*((208-94)/(6378.137+208)(6378.137+94))
added energy = 5.37*10^11
B. v1 = sqrt(GM/(R + h1))
v2 = sqrt(GM/(R + h2))
This is simply 1/2 m (v2)^2 - 1/2 m (v1)^2 = 0.5*1007*(GM/(R + h1) - GM/(R + h2))
0.5*1007*6.67*10^-11*5.98*10^24((208-94)/(6378.137+208)(6378.137+94))
5.37*10^11
C. This is simply mg(R + h2) - mg(R + h1)
mg(h2-h1) = 1007*9.8*(208-94) =1.125*10^7
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