Formulas
Aphelion, perihelion refer to farthest distance, and closest to the Sun. Apogee,
perigee refer to farthest distance and closest to the Earth. (peri is the closest)
a
e
p peri  aap
2
 p peri
aap
Finding the semi major axis. This is the
average distance from the orbiting body
How much it departs from a circle between o and 1
aap  p peri
 (1  e)
V p peri 
Velocity at peri (closest point), use correct
p peri
 (1  e)
Va ap 
Velocity at ap (farthest point), use correct
aap
Pperiod  a 3
2
Period using a in solar system only use if
orbiting the Sun.
Pperiod  2
Vescape
Vcircular
a3

2

rradius
Period of orbiting object in seconds, use
correct
The escape velocity from a body , use correct


rradius  hheight
p peri  a (1  e)
aap  a (1  e)
Velocity for object orbiting in a
circular orbit . Use correct
Closet distance the orbiting object comes to the
object being orbited in an elliptical orbit
Farthest distance the orbiting object comes to the
object being orbited in an elliptical orbit
Conversions
To change from km/sec to miles/hr
Km/sec (3600 sec/hr)(0.62137 miles/km= miles/hr
If the period is in seconds / 3600 for hours
To change from au to km multiply by 150,000,000 or 1.5 X 108 km
To change from miles to km multiply by 1.61 km/mile
e has no units.
There is a review of math with scientific notation on slide 15
** Be sure you use the  of the body that is being orbited ***
I did not always follow exactly significant figures, but I did not
use all the digits.
Answers may vary slightly depending upon how you round off
the decimals, and that’s ok.
The gravitational parameter
Body
Sun
Mercury
Venus
Earth
Mars
Jupiter
Saturn
Uranus
Neptune
Pluto
μ (km3s-2)
132,712,440,018
22,032
324,859
398,600
42,828
126,686,534
37,931,187
5,793,947
6,836,529
1,001
In Astrodynamics,the standard gravitational parameter of a celestial body is
the product of the gravitational constant
and the mass :
GM  
The units of the standard gravitational parameter are km3 s-2
**Be sure to use the right
for the object you are orbiting ! ****
Eccentricity of ellipse
eccentricity
e=
(
e)
0=circle, 1 = line
aphelion - perihelion
___________________________
aphelion + perihelion
e=0
e=0.98
Properties of Ellipses
semi-major axis = a
1/2 length of major axis
b
a
a = Semi-major axis
b = Semi-minor axis
a = aphelion
+ perihelion
_______________________
2
perihelion
aphelion
P2  a 3
Using the Kepler’sThird Law
P2  a3
P2  a3
if :
P measured in earth years, and a in AU.
A planet’s avg distance from the sun is 4 au,
what is the period of the planet ?
P  4 , p  64,
2
P
3
2
64, P  8 years
Astronomers use the metric system, whereas we are all more
familiar with the English system. I will use conversion so that you
can be more familiar with the answers.
I. The space shuttle is in a circular orbit 200 miles above the earth.
Find the period and velocity of the shuttle.
200miles(1.61 km/mile) =322 km above the Earth. The height of the
satellite in the problem must be the radius of the earth + height of
object. Radius of earth 6378km r+h = 6378+322= 6700 km
Vcircular 

rradius  hheight
398,000km3 / sec2
Vcircular 
 59.403km2 / sec2  7,707km / sec
6700km
(7,707km / sec)(3600sec/ hour )(0.62137mi / km)  17,240miles / hour
You can’t use Pperiod  a 3 because it circles the earth, not the Sun.
2
Pperiod  2
Pperiod
a3

a is the distance from center of Earth to the shuttle
(6700km)3
2
 2

2

755685.9sec
 2(3.141)(869.3)  5461.8sec
3
2
398,000km / sec
II. This problem covers a lot of formulas. An asteroid’s closest approach to the
sun is 2 au, and its farthest distance from the Sun is 4.5 au. Find a, the
eccentricity, distance at perihelion, distance at aphelion, period, velocity at
perihelion, and aphelion.
a
2
p peri  aap
e
aap  p peri
aap  p peri
2  4.5
a
 3.25au
2
4.5  2 2.5
e

 0.385
4.5  2 6.5
Find the perihelion, and aphelion distances.
p peri  a (1  e) = 3.25au (1 - 0.385) = (3.25)(.615) = 1.99 au
aap  a (1  e) = 3.25au (1+ 0.385) = 4.43 au
Pperiod  a
2
Find the period.
3
Pperiod  3.253
2
Pperiod  5.86 years
Perihelion, and aphelion must be changer to km, since  contains km . To change
multiply au by 150,000,000 km/au , or
1.5 X 108
V p peri 
 (1  e)
p peri
Vpperi  2.48km / sec
Vp peri
1.327 X 1011 (1  0.385)

1.99(1.5 X 108 )
Vp peri
18.38 X 1010

2.99 X 108
Vpperi  24.8km / sec
24.8 km/sec(3600)(.6213) = 55,469.7 miles/hour
Va ap 
Va 
 (1  e)
1.3267 x1011 (1  .385)
Va 
4.43(1.5 X 108 )
aap
0.816 X 1011
6, 645 X 108 )
8.16 X 1010
Va 
6.645 X 108 )
Va  3.504km / sec = 3.504(3600)(0.62137)= 7,838.8 mils/hour
III. An Earth satellite is in an elliptical orbit around the Earth; its
perigee is 160 km, and the apogee is 800 km. Find e, period, and
velocity at perigee, and apogee. Radius of the Earth = 6380 km
a ap
e
=800 +6380 = 7180km
aap  p peri
aap  p peri
e
p peri
= 160 + 6380 = 6540 km
7180  6540
 0.0466
7180  6540
V p peri 
Vp peri
 (1  e)
p peri
398,600km3 / sec2 (1.0466)

 8.17km / sec(3600)(.62137)  18, 277 miles / hour
6540km
Va ap 
 (1  e)
aap
398,600km3 / sec2 (1  0.0466)
Vaap 
 7.1km / sec(3600)(.62137)  15,890miles / hour
7180km
We need a
Pperiod  2
Pperiod
a
a3
p peri  aap
2
6540  7180
a
 6860km
2

(680km)3
3.228 X 1011
 2
 2(3.1415)
 5,654.2sec onds /3600 = 1.57 hrs
5
398,600km
3.986 X 10
I had to include my Halley’s comet problem. Halleys has a period of ~76 years
and e=0.967 Find the velocity at perihelion (close), and aphelion (far).
P 2  a3
762  a 3
a  17.94au
p perihelion  a (1  e) = 17.94(0.033) = .592 au (
7
1.5 X 108 ) = 8.88 X 10 km
aaphelion  a (1  e) = 17.94(1.967) = 35.45 au ( 1.5 X 108 ) = 5.385 X 109 km
V p peri 
 (1  e)
p peri
1.32712 X 1011 km3 / sec2 (1.967)
Vperihelion 
 54.343km / sec(3600)(.62137)  121,560miles / hour
7
8.88 X 10 km
Va ap 
 (1  e)
aap
1.32712 X 1011 km3 / sec2 (0.033)
Vaphelion 
 .9075km / sec(3600)(.62137)  2,030miles / hour
9
5.358 X 10 km
A Review of Scientific Notation Math
Multiplication:
•The digit terms are multiplied in the normal way and
the exponents are added. The end result is changed so
that there is only one nonzero digit to the left of the
decimal.
•Example: (3.4 x 106)(4.2 x 103) = (3.4)(4.2) x 10(6+3) =
14.28 x 109 = 1.4 x 1010
(to 2 significant figures)
•Example: (6.73 x 10-5)(2.91 x 102) = (6.73)(2.91) x 10(5+2) = 19.58 x 10-3 = 1.96 x 10-2
(to 3 significant figures)
Division:
•The digit terms are divided in the normal way and the
exponents are subtracted. The quotient is changed (if necessary)
so that there is only one nonzero digit to the left of the decimal.
•Example: (6.4 x 106)/(8.9 x 102) = (6.4)/(8.9) x 10(6-2) = 0.719 x 104
= 7.2 x 103
(to 2 significant figures)
•Example: (3.2 x 103)/(5.7 x 10-2) = (3.2)/(5.7) x 103-(-2) = 0.561 x
105 = 5.6 x 104
(to 2 significant figures)
Powers of Exponentials:
•The digit term is raised to the indicated power and the exponent is
multiplied by the number that indicates the power.
•Example: (2.4 x 104)3 = (2.4)3 x 10(4x3) = 13.824 x 1012 = 1.4 x 1013
(to 2 significant figures)
•Example: (6.53 x 10-3)2 = (6.53)2 x 10(-3)x2 = 42.64 x 10-6 = 4.26 x 10-5
(to 3 significant figures)
Roots of Exponentials:
•Change the exponent if necessary so that the number is divisible by the root.
Remember that taking the square root is the same as raising the number to
the one-half power.
•Example: