Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 539 km above the earth's surface, while that for satellite B is at a height of 876 km. How do you find the orbital speed for satellite A and satellite B?

Question

Two satellites are in circular orbits around the earth. The orbit for satellite A is at a height of 539 km above the earth's surface, while that for satellite B is at a height of 876 km.  How do you find the orbital speed for satellite A and satellite B?

Leah Angelini · Accepted Answer

#V_B=7408m/s#

To do this problem, you need the Earth's radius #R = 6371 km#

For the satellite to be in a stable orbit at a height, h, its centripetal acceleration #V^2/(R + h)# must equal the acceleration due to gravity at that distance from the center of the earth #g(R^2/(R + h)^2)#

#V^2/(R + h) = g(R^2/(R + h)^2)#

#V = sqrt(g(R^2/(R + h)))#

Regarding satellite A:

#V_A = sqrt(g(R^2/(R + h_A)))#

#V_A = sqrt(9.8 m/s^2((6371000 m)^2)/(6371000 m + 539000 m))#

#V_A = 18131 m/s#

Regarding satellite B:

#V_B = sqrt(g(R^2/(R + h_B)))#

#V_B = sqrt(9.8 m/s^2((6371000 m)^2)/(6371000 m + 876000 m))#

#V_B = 7408 m/s#

Leah Angelini · Answer

undefined The orbital speed of a satellite in a circular orbit around the Earth can be calculated using the formula:

$$ v = \sqrt{\frac{GM}{r}} $$

Where:
- $ v $ is the orbital speed (in meters per second).
- $ G $ is the gravitational constant ($ 6.674 \times 10^{-11} $ m³/kg/s²).
- $ M $ is the mass of the Earth ($ 5.972 \times 10^{24} $ kg).
- $ r $ is the radius of the orbit (Earth's radius plus the height of the orbit).

For satellite A:
$$ r_A = 539 \text{ km} + 6371 \text{ km} = 6910 \text{ km} $$

For satellite B:
$$ r_B = 876 \text{ km} + 6371 \text{ km} = 7247 \text{ km} $$

Now, using the formula:

$$ v_A = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6910000}} $$

$$ v_B = \sqrt{\frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{7247000}} $$

Calcula