How far (in meters) above the earth's surface will the acceleration of gravity be 27.0 % of what it is on the surface?
To determine the distance above the Earth's surface where the acceleration of gravity is 27.0% of what it is on the surface, we need to use the equation for gravitational acceleration:
g' = (1 - (d / R)) * g,
where g' is the acceleration of gravity at a given distance, d, above the surface of the Earth, R is the radius of the Earth, and g is the acceleration due to gravity at the surface.
We can rearrange the equation to solve for d:
d = (1 - (g' / g)) * R.
Given that g' is 27.0% of g, or g' = 0.27 * g, we can substitute this value into the equation:
d = (1 - (0.27 * g / g)) * R.
The ratio g/g cancels out, simplifying the equation to:
d = (1 - 0.27) * R.
Substituting the value of R, which is approximately 6,371 kilometers or 6,371,000 meters, we can calculate the distance:
d = 0.73 * 6,371,000.
Solving this equation, we find:
d = 4,650,430 meters.
Therefore, the distance above the Earth's surface where the acceleration of gravity is 27.0% of what it is on the surface is approximately 4,650,430 meters.