At what height above the earth is the acceleration due to gravity 12% of its value at the surface?
To determine the height above the Earth where the acceleration due to gravity is 12% of its value at the surface, we can use the concept of gravitational field strength. The gravitational field strength, denoted by g, is the acceleration experienced by an object due to the gravitational pull of a massive body.
At the surface of the Earth, the gravitational field strength is typically denoted as g₀, which is approximately 9.8 meters per second squared (m/s²). We need to find the height where the gravitational field strength is 12% of g₀.
Let's denote this height as h. To find h, we can use the following equation:
g = g₀ * (1 - (2h/r))
Where g is the gravitational field strength at height h, g₀ is the gravitational field strength at the surface, and r is the radius of the Earth.
In this case, we want g to be 12% of g₀:
g = 0.12 * g₀
Substituting this into the equation gives:
0.12 * g₀ = g₀ * (1 - (2h/r))
Now, we can simplify the equation by canceling out g₀:
0.12 = 1 - (2h/r)
To solve for h, we isolate it:
2h/r = 1 - 0.12
2h/r = 0.88
Now, let's solve for h:
2h = 0.88 * r
h = (0.88 * r) / 2
The value of the Earth's radius, r, is approximately 6,371 kilometers or 6,371,000 meters. Substituting this value into the equation will give us the height above the Earth's surface:
h = (0.88 * 6,371,000) / 2
h ≈ 2,795,720 meters
Therefore, the height above the Earth where the acceleration due to gravity is 12% of its value at the surface is approximately 2,795,720 meters.