A satellites are placed in a circular orbit that is 2.44 × 107 m above the surface of the earth. What is the magnitude of the acceleration due to gravity at this distance?
A better name is magnitude of gravitational field force.
g= 9.8*(re/(re+2.44E7))^2 and the units are Newton/kg, of if you prefer, m/s^2
make certain radius earth re is in meters.
To find the magnitude of the acceleration due to gravity at a certain distance from Earth's surface, we can use the formula for the acceleration due to gravity:
a = GM / r^2
Where:
a is the acceleration due to gravity
G is the gravitational constant (approximately 6.674 × 10^-11 m^3/kg/s^2)
M is the mass of Earth (approximately 5.972 × 10^24 kg)
r is the distance from the center of Earth to the satellite's orbit (2.44 × 10^7 m + radius of Earth)
First, we need to calculate the actual distance from the center of Earth to the satellite's orbit. We add the distance above the surface (2.44 × 10^7 m) to the radius of Earth (approximately 6.371 × 10^6 m):
Distance = 2.44 × 10^7 m + 6.371 × 10^6 m
Next, we substitute the values into the formula:
a = (6.674 × 10^-11 m^3/kg/s^2) × (5.972 × 10^24 kg) / (Distance)^2
After simplifying the equation and performing the calculation, the magnitude of the acceleration due to gravity at this distance is obtained.