What is the acceleration due to gravity at 7220km above the Earth's surface? Take the radius of the Earth to be 6.37e6m and the mass of the Earth to be 5.972e24kg.
g=G•M/(R+h)²,
the gravitational constant
G =6.67•10^-11 N•m²/kg²,
Earth’s mass is M = 5.972•10^24 kg,
Earth’s radius is R = 6.37 •10^6 m.
h=7220000 m
1.69 m/s^2
To find the acceleration due to gravity at a certain height above the Earth's surface, we can use the formula:
g = (G * M) / (r + h)^2
Where:
g is the acceleration due to gravity,
G is the universal gravitational constant (approximately 6.67e-11 m^3/kg/s^2),
M is the mass of the Earth, and
r is the radius of the Earth.
Given:
r = 6.37e6 m (radius of the Earth)
M = 5.972e24 kg (mass of the Earth)
h = 7220 km = 7220000 m (height above the Earth's surface)
Now, we can substitute these values into the formula and calculate the acceleration due to gravity:
g = (G * M) / (r + h)^2
= (6.67e-11 * 5.972e24) / (6.37e6 + 7220000)^2
= (3.972e14) / (1.3605e7)^2
= (3.972e14) / (1.8527e14)
≈ 0.214 m/s^2 (approximately)
Therefore, the acceleration due to gravity at a height of 7220 km above the Earth's surface is approximately 0.214 m/s^2.