Geosynchronous communications satellites are placed in a circular orbit that is 3.59 107
m
above the surface of the earth. What is the magnitude of the acceleration due to gravity at this
distance?
g at r=6.37 * 10^6 m is 9.8
3.59*10^7 = 5.63r, g = 9.8/5.63^2 = 0.31 m/s^2
To find the magnitude of the acceleration due to gravity at a certain distance from the surface of the Earth, you can use the formula for gravitational acceleration:
g = G * (M/R^2)
Where:
g is the acceleration due to gravity,
G is the gravitational constant (approximately 6.67 × 10^-11 N m²/kg²),
M is the mass of the Earth (approximately 5.98 × 10^24 kg),
R is the distance between the center of the Earth and the satellite (3.59 × 10^7 m).
Substituting the values into the formula:
g = (6.67 × 10^-11 N m²/kg²) * (5.98 × 10^24 kg) / (3.59 × 10^7 m)^2
Using a calculator to evaluate this expression, we find:
g ≈ 9.80 m/s²
Therefore, the magnitude of the acceleration due to gravity at a distance of 3.59 × 10^7 m above the surface of the Earth is approximately 9.80 m/s².