At what height above Earth's surface is the gravitational acceleration reduced from its sea-level value by 0.40%?
To determine the height above Earth's surface at which the gravitational acceleration is reduced by 0.40%, we can use the concept of the variation of acceleration due to gravity with height.
The acceleration due to gravity, denoted as "g," decreases with increasing height from the Earth's surface. The rate at which the acceleration due to gravity decreases is determined by the universal law of gravitation.
The formula to calculate the variation of gravity with height is as follows:
g' = g * (1 - (2 * h / R))
Where:
g' is the gravitational acceleration at height h
g is the gravitational acceleration at the Earth's surface
h is the height above Earth's surface
R is the radius of the Earth
In this case, we need to find the height at which the gravitational acceleration is reduced by 0.40%. We can set up the equation as:
0.994 * g = g * (1 - (2 * h / R))
Simplifying the equation, we have:
0.994 = 1 - (2 * h / R)
Rearranging the equation to solve for h:
2 * h / R = 1 - 0.994
2 * h / R = 0.006
h / R = 0.006 / 2
h / R = 0.003
Multiplying both sides by R:
h = 0.003 * R
To find the specific height, we need to know the value of R, the radius of the Earth, which is approximately 6,371 kilometers.
h = 0.003 * 6,371 kilometers
Calculating the result:
h ≈ 19.11 kilometers
Therefore, at approximately 19.11 kilometers above the Earth's surface, the gravitational acceleration is reduced from its sea-level value by 0.40%.