Obtain the expression for kinetic energy, potential energy ,mechanical energy of artificial satellite moving around the earth
Answers
The following four statements about circular orbits are equivalent. Derive any one of them from first principles.
Negative kinetic energy equals half the potential energy (−K = ½U).
Potential energy equals twice the total energy (U = 2E).
Total energy equals negative kinetic energy (E = −K).
Twice the kinetic energy plus the potential energy equals zero (2K + U = 0).
This is a key relationship for a larger problem in orbital mechanics known as the virial theorem.
solution
Circular orbits arise whenever the gravitational force on a satellite equals the centripetal force needed to move it with uniform circular motion.
Fc = Fg
mv2 = GMm
rp rp2
v2 = Gm1
r
Substitute this expression into the formula for kinetic energy.
K = 1 m2v2
2
K = 1 m2 ⎛
⎝ Gm1 ⎞
⎠
2 r
K = 1 Gm1m2
2 r
Note how similar this new formula is to the gravitational potential energy formula.
K = + 1 Gm1m2
2 r
Ug = − Gm1m2
r
K = − 1 Ug
2
The kinetic energy of a satellite in a circular orbit is half its gravitational energy and is positive instead of negative. When U and K are combined, their total is half the gravitational potential energy.
E = K + Ug
E = − 1 Ug + Ug
2
E = 1 Ug
2
E = − Gm1m2
2r
The gravitational field of a planet or star is like a well. The kinetic energy of a satellite in orbit or a person on the surface sets the limit as to how high they can "climb out of the pit". A satellite in a circular orbit is halfway out of the pit (or halfway in, for you pessimists).
practice problem 2
Determine the minimum energy required to place a large (five metric ton) telecommunications satellite in a geostationary orbit.
solution
Start by determining the radius of a geosynchronous orbit. There are several ways to do this (which includes looking it up somewhere), but the traditional way is to start from the principle that the centripetal force on a satellite in a circular orbit is provided by the gravitational force of the Earth on the satellite. Combine this with the formula for the speed of an object in uniform circular motion. The algebra is somewhat tedious and has been condensed in the derivation below.
Fc = mv2 = Gm1m2 = Fg
r r2
v = Δs = 2πr
Δt T
r = ⎛
⎝ GmT2 ⎞⅓
⎠
4π2
rf = ⎛
⎝ (6.67 × 10−11 Nm2/kg2)(5.97 × 1024 kg)(24 × 60 × 60 s)2 ⎞⅓
⎠
4π2
rf = 4.223 × 107 m (geostationary orbit)
Next, use the relationship derived in the previous problem to determine the total energy of the satellite in orbit. This will be the final energy of the system.
Ef = Kf + Uf = Uf
2
Ef = − Gm1m2
2rf
Ef = − (6.67 × 10−11 Nm2/kg2)(5.97 × 1024 kg)(5,000 kg)
2(4.225 × 107 m)
Ef = −2.357 × 1010 J
Ef = −23.57 GJ (geostationary orbit)
To satisfy the minimum energy requirements of this problem the satellite should be launched from someplace on the equator where the speed of rotation (and thus the kinetic energy) is a maximum.
vi = Δs = 2πr
Δt T
vi = 2π(6.37 × 106 m)
(24 × 60 × 60 s)
vi = 463.2 m/s (on the equator)
The initial energy of the satellite is the gravitational potential energy it has on the Earth's surface plus the kinetic energy it has due to the Earth's rotation. (Remember, gravitational potential energy is negative.)
Ei = Ki + Ui
Ei = 1 mvi2 − Gm1m2
2 ri
Ei =
1 (5,000 kg)(463.2 m/s)2
2
− (6.67 × 10−11 Nm2/kg2)(5.97 × 1024 kg)(5,000 kg)
(6.37 × 106 m)
Ei = −3.120 × 1011 J
Ei = −312.0 GJ (on the equator)
Subtract the initial and final energies to finish the problem.
ΔE = Ef − Ei
ΔE = (−23.57 GJ) − (−312.0 GJ)
ΔE = 288 GJ