Dr. Neal, WKU
MATH 117
Orbital Motion Around Earth
According to Newton’s Law of Gravitation, the acceleration due to gravity g of any
celestial body is inversely proportional to the square of the distance to the center of
C
mass of the body. For a planet, g = 2 , where C is a constant that depends on the
r
planet, and r is the distance to the center of the planet. The gravity decreases as r
increases.
For an object in a circular orbital motion around a planet, such as a satellite or Space
v2
Shuttle, the outward centripetal acceleration due to its motion is given by a =
. In
r
order for the object to stay in orbit, the centripetal acceleration a must be in
equilibrium with the gravity of the planet g . Thus, we must have a = g , or
v2 C
= 2 .
r
r
To obtain the speed necessary to stay in orbit, we solve for v :
v=
C
r
The Orbital Constant C
C
Because g = 2 , if we know the gravity on the surface and the radius of the planet, then
r
we can solve for C = g r 2 . For our purposes, we will want all units to be in miles and
hours. The constant C will be in (miles)3/(hrs)2. To obtain these units, we use
C = g r2 = g
2
1 mile
ft
2
2
2 sec
mile
!
!
!
r
.
3600
5280 ft
sec 2
hr 2
For Earth, g ≈ 32.174 ft / sec 2 and r ≈ 3963.2 miles; thus,
C ≈ 32.174 ! (3963. 2)2 !
1
miles 3
! 36002 ≈ 1.24 ! 1012
.
5280
hr 2
3
1 2 miles
2
C = 1.24 ! 10
hr
Dr. Neal, WKU
Circular Orbit
In order to stay in circular orbit above Earth, the velocity must satisfy v =
C
=
r
1.24 ! 1012
, where r is the distance to the center of Earth. Using r = 3963.2 + altitude,
r
1. 24 ! 1012
the orbital velocity then must satisfy v =
mph.
(3963.2 + a)
Here is a summary of the equations we need:
The radius of the Earth is about 3963.2 miles; g ≈ 32.174 ft / sec 2
Earth rotates once in 23h 56m 4.1 s ≈ 23.9345 hrs.
(a) Orbital Velocity
v=
(b) Angular Velocity of Earth
(How fast the angle of rotation
moves as Earth rotates.)
1.24 ! 1012
mph
d
2"
radians
!E = #
56
4.1 &
hr
% 23 +
+
(
$
'
60 3600
radians
) 0.262516
hr
where d is the distance to center of Earth
or
v=
1. 24 ! 1012
mph
(3963.2 + a)
Can multiply by
where a is the altitude above Earth
(c) Velocity from Angular Velocity
(For ! in rad per hr)
v =!r
and
! E " 15. 041º per hour
(d) Orbital Angular Velocity of Satellite
!S =
v
!=
r
v
v
radians
=
d (3963.2 + a)
hr
where v is found from Part (a)
For Earth’s spin, ! E " 0. 262516 rad/hr
and r ! 3963. 2 miles at the equator
So the velocity of the equator spin is
180º
to get
!
Multiply by
180º
to get deg per hr
!
v = ! r = 0. 262516 " 3963.2 # 1040. 4 mph
Centripetal Acceleration
acc =
v2 C
1.24 ! 1012 miles
= 2 =
r
r
(3963.2 + a)2 hr2
" 5280
ft %
$!
to put in 2 '
2
# 3600
s &
Dr. Neal, WKU
Example. Suppose the Space Shuttle is 220 miles above Earth orbiting in the plane of
the equator.
(a)
(b)
(c)
(d)
(e)
What speed is needed to maintain orbit at 220 miles altitude?
What is the centripetal acceleration (in ft/s2 ) at this altitude? What pct. of g ?
What is the Shuttle’s angular velocity?
How long will it take the Shuttle to complete one 360º orbit?
How long does it take for the shuttle returns to its starting point above the equator?
Solution. (a) To stay in orbit, v =
1. 24 ! 1012
=
(3963.2 + a)
1. 24 ! 1012
≈ 17,216.96 mph.
(3963.2 + 220)
1. 24 ! 1012 miles
C
ft
v2
2
= 2 =
2
2 ( ! 5280 /3600 ) ≈ 28.87
2 (divide by
r
(3963. 2 + 220) hr
r
sec
met
3.28 to get ≈ 8.8
2 ). These values give the gravity pull of Earth at the 220 mile
sec
altitude, which is 28.87/32.174 = 0.8973, or about 89.73% of g.
(b) acc =
(c) ! =
v
v
17216.96
=
=
≈ 4.11574 (rad)/hr (which is about 235.81º/hr)
r (3963.2 + a)
(3963. 2 + 220)
(d) The time t for one 360º = 2π radian revolution of the Shuttle, is given by
4.11574 t = 2π
! t=
2!
≈ 1.5266 hrs ≈ 1 hr, 31 min, 36 sec.
4.11574
(e) The angular velocity function of the Shuttle at this altitude is 4.11574 t . The angular
2!
t
velocity function of Earth is "
56
4.1 % t ≈ 0.262516
$ 23 +
+
'
#
60 3600 &
For shuttle to lap Earth, we must find the time t such that
4.11574 t = 0.262516 t + 2π.
That is, the shuttle’s angle must equal the Earth’s angle plus one revolution. Solving for
t we have 3.853224 t = 2π and t ≈ 1.63 hrs ≈ 1 hr, 37 min, 50 sec.
Dr. Neal, WKU
Exercises
1. A communications satellite is 18,200 miles above the Earth orbiting in the plane of
the equator.
(a)
(b)
(c)
(d)
(e)
What speed must it have in order to maintain orbit?
What is its centripetal acceleration in ft/s2?
What is its angular velocity?
How long will it take to complete one 360º orbit?
How long until it returns to its starting point above the equator?
2. A probe is 75 miles above Mars orbiting in the plane of the equator. (The radius of
Mars is 2108.56 miles, and its orbital constant is C ≈ 1.33 ! 1011 (miles)3/(hrs)2 . Mars
rotates in 24h 37m 22.662s.
(a) What speed must the probe have in order to maintain orbit?
(b) What is its angular velocity?
(c) How long until the probe returns to its starting point above Mars’ equator?
Dr. Neal, WKU
Solutions
1. (a) v =
(b) acc =
1.24 ! 1012
≈ 7,479.88 mph.
3963.2 + 18, 200
1.24 ! 1012
miles
ft
2
.
2 ( ! 5280 /3600 ) ≈ 1.03
2
(3963.2 + 18, 200) hr
sec 2
(c) ! = v / r ≈ 7,479.88 / (3963.2 + 18,200) ≈ 0.33749 (rad)/hr. (≈ 19.337 deg per hr)
(d) Set 0.33749 t = 2π ! t ≈ 18.6174 hrs ≈ 18 hr, 37 min, 3 sec
(e) The Earth’s angular velocity function is 0.262516 t . For the satellite to lap Earth,
then 0.33749 t = 0.262516 t + 2π. So 0.074974 t = 2π and t ≈ 83.8 hrs
2. (a) v =
1.33 "1011
≈ 7804.47 mph.
2108.56 + 75
(b) ! = v / r ≈ 7804.47 / (2108.56 + 75) ≈ 3.5742 (rad)/hr.
So the angular velocity function is 3.5742 t .
!
(c) For Mars’ equator, the angular velocity function is
2!
"
37 22.662 % t ≈ 0.255176 (rad)/hr.
$ 24 +
+
'
#
60
3600 &
Thus, set 3.5742 t = 0.255176 t + 2π ! t ≈ 1.893 hrs ≈ 1 hr, 53 min, 35 sec.
!