Orbital Mechanics II
practice problem 1
- 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).
Circular orbits arise whenever the gravitational force on a satellite equals the centripetal force needed to move it with uniform circular motion.
Fc = Fg
Substitute this expression into the formula for kinetic energy.
K = ½m2v2
|K = ½m2||⎛
|K = ½||Gm1m2|
Note how similar this new formula is to the gravitational potential energy formula.
|K = + ½||Gm1m2|
|Ug = −||Gm1m2|
K = −½Ug
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 = −½Ug + Ug
E = ½Ug
|E = −||Gm1m2|
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 well. A satellite in a circular orbit is halfway out (or halfway in, for you pessimists).
practice problem 2
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|
|(6.67 × 10−11 N m2/kg2)
rf = 4.223 × 107 m (geostationary orbit)
Next, use the virial theorem to determine the total energy of the satellite in orbit. This will be the final energy of the system.
|Ef = Kf + Uf =||Uf|
|Ef = −||Gm1m2|
|Ef = −||(6.67 × 10−11 N m2/kg2)
|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 =||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|
Ei = −3.120 × 1011 J
Ei = −312.0 GJ (on the equator)
Subtract the initial and final energies to finish the problem.
|∆E = Ef − Ei
∆E = (−23.57 GJ) − (−312.0 GJ)
∆E = 288 GJ
practice problem 3
- The satellite is initially in an elliptical orbit as shown in the diagram to the right. At perigee (the point of closest approach) the distance from the center of the satellite to the center of the Earth is rp and the speed of the satellite is vp. At apogee (the point when it is furthest from the Earth) the distance from the center of the satellite to the center of the Earth is ra. Determine va, the speed at apogee.
- As the satellite reaches perigee, its speed is changed abruptly so that the satellite enters a circular orbit of radius rp and speed v as shown in the diagram to the right. How much work and what impulse was applied to the satellite to change its orbit?
There are two ways to solve the first half of this problem: using conservation of energy and conservation of angular momentum. The symbolic answers look different, but the numeric answers they generate are always the same. Establishing this fact algebraically is quite challenging, however.
The total energy of a satellite is just the sum of its gravitational potential and kinetic energies. Assuming that mechanical energy is conserved (which it is for the most part in the vacuum of space), the sum of the kinetic and potential energies of the satellite would remain constant. Set this sum at apogee equal to the sum at perigee and solve for the speed at apogee.
Ka + Ua = Kp + Up 1 mva2 − Gm1m2 = 1 mvp2 − Gm1m2 2 ra 2 rp va2 − vp2 = 2GM ⎛
1 − 1 ⎞
ra rp va = ⎡
vp2 + 2GM ⎛
1 − 1 ⎞
For the same reason that mechanical energy is conserved angular momentum is also conserved as the satellite moves from perigee to apogee. The speed at apogee can be determined quite simply from this principle. Set the angular momentum of the satellite at apogee equal to the angular momentum at perigee and solve for va.
La = Lp mvara = mvprp va = rp vp ra
The centripetal force on a satellite in a circular orbit is provided by gravity. Set the two equations equal to one another and solve for v. The equation so derived will be used twice.
Fc = Fg mv2 = Gm1m2 rp rp2 v = √ GM rp
For the second and third parts of this problem, apply the impulse-momentum theorem…
J = Δp = mΔv = m(v − vp)
J = m ⎛
GM − vp ⎞½
and the work-energy theorem.
W = ΔK = 1 mv2 − 1 mvp2 2 2 W = 1 m ⎛
− 1 mvp2 2 rp 2 W = m ⎛
GM − vp2 ⎞
practice problem 4
- from the Sun and Earth in meters
- from the Sun as multiples of the Earth's orbital radius (au)
- from the Earth as multiples of the Moon's orbital radius