at what altitude the value of g becomes 3/4 of its value on the surface of earth ?
To determine the altitude at which the value of g becomes 3/4 of its value on the surface of the Earth, we need to use the formula for the gravitational acceleration:
g = GM/R^2
where:
g is the gravitational acceleration
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
M is the mass of the Earth (approximately 5.97 × 10^24 kg)
R is the distance from the center of the Earth to the altitude we want to find
First, we need to calculate the value of g on the surface of the Earth. Plugging in the known values:
g(surface) = (6.67430 × 10^-11) * (5.97 × 10^24) / (6,371,000 meters)^2
g(surface) ≈ 9.81 m/s^2
Now we can set up an equation to solve for the altitude (R):
3/4 * g(surface) = (6.67430 × 10^-11) * (5.97 × 10^24) / R^2
Multiplying both sides by R^2:
(3/4) * g(surface) * R^2 = (6.67430 × 10^-11) * (5.97 × 10^24)
R^2 = [(6.67430 × 10^-11) * (5.97 × 10^24)] / [(3/4) * g(surface)]
R^2 ≈ (1.991 × 10^14) / (7.356 × 9.81)
R^2 ≈ 2.628 × 10^11
Taking the square root of both sides:
R ≈ √(2.628 × 10^11)
R ≈ 1.619 × 10^5 meters
Therefore, the altitude at which the value of g becomes 3/4 of its value on the surface of the Earth is approximately 161,900 meters (or 161.9 kilometers) above the Earth's surface.