How far (in meters) above the earth's surface will the acceleration of gravity be 34.0 % of what it is on the surface?
To find out how far above the Earth's surface the acceleration of gravity will be 34.0% of what it is on the surface, we can use the fact that the acceleration due to gravity decreases with increasing distance from the Earth's surface.
The formula for the acceleration due to gravity at a certain distance above the Earth's surface is given by:
g' = g * (R / (R + h))^2
Where:
- g is the acceleration due to gravity on the Earth's surface (approximately 9.8 m/s^2)
- g' is the acceleration due to gravity at the given distance above the surface
- R is the radius of the Earth (approximately 6,371,000 meters)
- h is the height (or distance above the Earth's surface) at which we want to find the acceleration due to gravity
According to the problem, we want to find the height (h) at which the acceleration of gravity is 34.0% of what it is on the surface. Therefore, g' = 0.34 * g.
Substituting these values into the formula, we can solve for h:
0.34 * g = g * (R / (R + h))^2
Dividing both sides of the equation by g and taking the square root, we get:
sqrt(0.34) = R / (R + h)
Simplifying further, we have:
0.34 = (R^2) / (R + h)^2
Cross multiplying the equation:
0.34 * (R + h)^2 = R^2
Expanding and rearranging the terms:
0.34 * (R^2 + 2Rh + h^2) = R^2
Multiplying out 0.34:
0.34 * R^2 + 0.68 * Rh + 0.34 * h^2 = R^2
Isolating the terms with h:
0.34 * h^2 + 0.68 * Rh + (0.34 - 1) * R^2 = 0
Simplifying further:
0.34 * h^2 + 0.68 * Rh - 0.66 * R^2 = 0
This is a quadratic equation in h. We can use the quadratic formula to solve for h:
h = (-b ± sqrt(b^2 - 4ac)) / 2a
In this equation, a = 0.34, b = 0.68 * R, and c = -0.66 * R^2.
Plugging in the values, we get:
h = (-(0.68 * R) ± sqrt((0.68 * R)^2 - 4 * 0.34 * (-0.66 * R^2))) / (2 * 0.34)
Solving this equation will give us the height (h) above the Earth's surface where the acceleration due to gravity is 34.0% of what it is on the surface.