http://www.ck12.org Chapter 9. Newton’s Universal Law of Gravity
Answer: With this height above the surface of the Earth, the satellite has a distance from the center of the Earth of
- 5 Re. The force of gravity is then
F=ma=Gmme
( 1. 5 Re)^2=
1
1. 52
Gmme
R^2 e=
1
1. 52
mga=1
1. 52
g=1
1. 52
( 9. 81 m/s^2 ) = 4. 36 m/s^2- If the satellite is in a circular orbit at this height, what is its velocity?
Answer:In this case, the acceleration due to gravity is the centripetal acceleration.
a=
v^2
r→v^2 =ar→v=√
arv=√
( 4 .36m/s^2 )(6370km)v=√
( 4 .36m/s^2 )( 6. 37 × 106 m)
v= 5. 27 × 103 m/sOrbital Altitudes
Satellite orbits are typically split into three groups: low, medium, and high. Low orbits extend to a height of 2000
km, while medium go up to approximately 20,000 km.
FIGURE 9.10
Illustrative Example 1
The satellite inFigure9.11 is in orbit at a heighthof 250 km above the surface of the Earth. The radius of the Earth
varies. For this, useRe=6370km.