Orbital velocity: the instantaneous velocity of an object moving in an elliptical orbit, due to the influence of gravity

Formula: v2 = GM(2/r - 1/a)

where G = 6.67 x 10-11 N m2 / kg2,
M is the mass of the planet (or object to be orbited),
r is the radial distance of the orbiting object from the center of the planet (or object to be orbited) at a given moment
a is the semi-major axis of the elliptical orbit


Calculate the instantaneous velocity of the Earth when it is closest and furthest from the Sun:

    at perihelion, the Earth is at 0.983 AU
    at aphelion, the Earth is at 1.017 AU
    1 AU = 150 x 106 km = 1.5 x 1011 m
    the mass of the Sun is 2 x 1030 kg

    vperihelion= [(6.67 x 10-11 x 2 x 1030)*(2/{0.983*1.5 x 1011} - 1/{1*1.5 x 1011})]0.5
    vperihelion= [(6.67 x 10-11 x 2 x 1030 /1.5 x 1011)*(2/0.983 - 1/1)]0.5
    vperihelion= [(8.87 x 108)*(2/0.983 - 1/1)]0.5
     vperihelion= [(8.87 x 108)(1.017)]0.5
     vperihelion= 3.00 x 104 m/s = 30.0 km/s
 

     vaphelion= [(8.87 x 108)*(2/1.017 - 1/1)]0.5
     vaphelion= [(8.87 x 108)(0.966)]0.5
     vaphelion= 2.93 x 104 m/s = 29.3 km/s

What will happen to a satellite in a circular orbit if we fire the `booster' rockets briefly?
 

assume the satellite is 200 km above the surface of Mars, in a circular orbit
assume we increase the rocket's speed, instantaneously, by 1 km/sec
 

then, initially
                        ainitial = r = Rmars+200 km = 3398 km + 200 km
                        ainitial = 3598 km
                        ainitial = 3.60 x 106 m
also, since the initial orbit is circular
                         rinitial = ainitial = 3.60 x 106 m
and
               vinitial = (GMmars/a)0.5
               vinitial = (6.67 x 10-11 x 6.42 x 1023 kg / 3.60 x 106 m)0.5
               vinitial = 3.45 x 103 m/s
               vinitial = 3.45 km/s

finally, using Kepler's third law, we can find the orbital period:

      a3/ p2 = G (M1 + M2) / 4 pi2
      p2 = 4 pi2 a3/  G (M1 + M2)
      p2 = 4 pi2 * (3.60 x 106)3 / {6.67 x 10-11 * (6.42 x 1023 kg + 2000 kg)}
     p2 = 1.84 x 1021 / {6.67 x 10-11 * 6.42 x 1023 }
     p2 = 1.84 x 1021 / {6.67 x 10-11 * 6.42 x 1023 }
     p2 = 4.30 x 107
     p = 6.56 x 103 seconds
     p = 109 minutes
 
 
 
 
 

after we fire the rockets, the rocket is at the same position but moving faster:
               vnew = vinitial + 1 km/s = 4.45 km/s
               rnew = rinitial
               anew = ?????

Now, the spacecraft has too much energy to have the same circular orbit.  Since it has more energy, it will rise up to a higher elevation orbit.  The point at which the rockets were fired is the only point in the old orbit which is also a point in the new orbit.  (In orbital dynamics, the object always returns to the "scene of the crime.")
so using the velocity equation, we can solve for the new orbital semi-major axis:

v2 = G Mmars(2/r - 1/a)

v2 / G Mmars = 2/r - 1/a

1/a = 2/r - v2 / GMmars
1/a = (2 G Mmars - r v2 ) / r G Mmars

a = r G Mmars / (2 G Mmars - r v2 )

a = 3.60 x 106  x 6.67 x 10-11 x 6.42 x 1023  / (2 x 6.67 x 10-11 x 6.42 x 1023  - 3.60 x 106 x [4.45 x 103 m/s]2)

a = 1.54 x 1020 /(8.56 x 1013 - 7.13 x 1013)
a = 1.08 x 107 m
a = 1.08 x 104 km
a = 10,800 km

What else do we know based on this?  We know that

rperi = 3.60 x 103 km
a = 10,800 km
and
rperi = a (1- e)
hence,
3,600 = 10,800 km (1- e)
3600/10,800 = 1 - e
0.33 = 1 - e
thus,
e = 0.67

and

raph = a (1+ e)
raph = 10,800 km (1 + 0.67)
raph = 18,000 km

so, the rocket is now in an extremely elliptical orbit that takes it as far from the surface of the planet as 18,000-3398 km = 15,702 km = 4.6 Rmars and as close to the surface of the planet as 3598 - 3398 km = 200 km,
i.e., skimming through the top of the atmosphere.

 again, using Kepler's third law, we can find the orbital period:

     a3/ p2 = G (M1 + M2) / 4 pi2
     p2 = 4 pi2 * (10.80 x 106)3 / {6.67 x 10-11 * (6.42 x 1023 kg + 2000 kg)}
     p2 = 4.97 x 1022 / {6.67 x 10-11 * 6.42 x 1023 }
     p2 = 1.16 x 109
     p = 3.41 x 104 seconds
     p = 568 minutes
     p = 9.5 hours
 
 

NASA example:

Mars '01 Odyssey Mission: ending aerobraking:

January 11, 2002

"Flight controllers for NASA's Mars Odyssey spacecraft sent commands overnight to raise the spacecraft up out of the atmosphere and conclude the aerobraking phase of the mission.  At 12:18 a.m. Pacific time Jan. 11,  Odyssey fired its small thrusters for 244 seconds, changing its speed by 20 meters per second (45 miles per hour) and raising its orbit by 85 kilometers (53 miles). The closest point in Odyssey's orbit, called the periapsis, is now 201 kilometers (125 miles) above the surface of Mars. The farthest point in the orbit, called the apoapsis, is at an altitude of 500 kilometers (311 miles). During the next few weeks, flight controllers will refine the orbit until the spacecraft reaches its final mapping altitude, a 400-kilometer (249-mile) circular orbit.