**Formula: **v^{2} = GM(2/r - 1/a)

where G = 6.67 x 10^{-11}N m^{2}/ kg^{2},

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 10^{6} km = 1.5
x 10^{11
}m

the mass of the Sun is 2 x 10^{30}
kg

v_{perihelion}= [(6.67 x 10^{-11}
x 2 x 10^{30})*(2/{0.983*1.5 x 10^{11}} - 1/{1*1.5 x 10^{11}})]^{0.5}

v_{perihelion}= [(6.67 x 10^{-11}
x 2 x 10^{30 }/1.5 x 10^{11})*(2/0.983 - 1/1)]^{0.5}

v_{perihelion}= [(8.87 x 10^{8})*(2/0.983
- 1/1)]^{0.5}

v_{perihelion}= [(8.87
x 10^{8})(1.017)]^{0.5}
^{ }v_{perihelion}=
3.00 x 10^{4} m/s = 30.0 km/s

v_{aphelion}= [(8.87
x 10^{8})*(2/1.017 - 1/1)]^{0.5}

v_{aphelion}= [(8.87
x 10^{8})(0.966)]^{0.5}
^{ }v_{aphelion}=
2.93 x 10^{4} 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

a_{initial}= r = R_{mars}+200 km = 3398 km + 200 km

a_{initial }= 3598 km

a_{initial }= 3.60 x 10^{6}m

also, since the initial orbit is circular

r_{initial }= a_{initial }= 3.60 x 10^{6}m

and

v_{initial}= (GM_{mars}/a)^{0.5}

v_{initial}= (6.67 x 10^{-11}x 6.42 x 10^{23}kg / 3.60 x 10^{6}m)^{0.5}

v_{initial}= 3.45 x 10^{3}m/s

v_{initial}= 3.45 km/sfinally, using Kepler's third law, we can find the orbital period:

a

^{3}/ p^{2 }= G (M_{1}+ M_{2}) / 4 pi^{2}

p^{2 }= 4 pi^{2 }a^{3}/ G (M_{1}+ M_{2})

p^{2 }= 4 pi^{2 }* (3.60 x 10^{6})^{3}/ {6.67 x 10^{-11}* (6.42 x 10^{23}kg + 2000 kg)}

p^{2 }= 1.84 x 10^{21}/ {6.67 x 10^{-11}* 6.42 x 10^{23}}

p^{2 }= 1.84 x 10^{21}/ {6.67 x 10^{-11}* 6.42 x 10^{23}}

p^{2 }= 4.30 x 10^{7}

p^{ }= 6.56 x 10^{3}seconds

p = 109 minutes

after we fire the rockets, the rocket is at the same position but moving faster:

v_{new}= v_{initial}+ 1 km/s = 4.45 km/s

r_{new}= r_{initial}

a_{new}= ?????

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:v

^{2}= G M_{mars}(2/r - 1/a)v

^{2}/ G M_{mars}= 2/r - 1/a1/a = 2/r - v

^{2}/ GM_{mars}

1/a = (2 G M_{mars}- r v^{2 }) / r G M_{mars}a = r G M

_{mars}/ (2 G M_{mars}- r v^{2})a = 3.60 x 10

^{6}x 6.67 x 10^{-11}x 6.42 x 10^{23}/ (2 x 6.67 x 10^{-11}x 6.42 x 10^{23}- 3.60 x 10^{6}x [4.45 x 10^{3}m/s]^{2})a = 1.54 x 10

^{20}/(8.56 x 10^{13}- 7.13 x 10^{13})

a = 1.08 x 10^{7}m

a = 1.08 x 10^{4}km

a = 10,800 kmWhat else do we know based on this? We know that

ragain, using Kepler's third law, we can find the orbital period:_{peri }= 3.60 x 10^{3}km

a = 10,800 km

and

r_{peri }= a (1- e)

hence,

3,600 = 10,800 km (1- e)

3600/10,800 = 1 - e

0.33 = 1 - e

thus,

e = 0.67and

r

_{aph }= a (1+ e)

r_{aph }= 10,800 km (1 + 0.67)

r_{aph }= 18,000 kmso, 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 R

_{mars}and as close to the surface of the planet as 3598 - 3398 km = 200 km,

i.e., skimming through the top of the atmosphere.

a^{3}/ p^{2 }=
G (M_{1} + M_{2}) / 4 pi^{2}

p^{2 }= 4 pi^{2 }*
(10.80 x 10^{6})^{3} / {6.67 x 10^{-11} * (6.42
x 10^{23} kg + 2000 kg)}

p^{2 }= 4.97 x 10^{22}
/ {6.67 x 10^{-11} * 6.42 x 10^{23} }

p^{2 }= 1.16 x 10^{9}

p^{ }= 3.41 x 10^{4}
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.