Calculate the effective value of g, the acceleration of gravity (in meters/second^2), at 5100 km above the earth's surface.

The effective value of g is inversely proportional to the SQUARE of the distance from the center of the earth.

The radius of the Earth is 6378 km. At 5100 km altitude, you are 11,478 km from the center.

g will lowered by a factor (6378/11478)^2
Credit to "drwl"

To calculate the effective value of g (acceleration due to gravity) at a given height above the Earth's surface, we need to use the formula:

g' = g * (R / (R + h))^2

Where:
g = acceleration due to gravity at the Earth's surface (9.8 m/s^2)
R = radius of the Earth (6371 km or 6,371,000 meters)
h = height above the Earth's surface (in meters)

In this case, the height is given in kilometers, so we need to convert it to meters. Since 1 km = 1000 meters, the height of 5100 km is equal to 5100 * 1000 = 5,100,000 meters.

Now we can plug these values into the formula:

g' = 9.8 * (6,371,000 / (6,371,000 + 5,100,000))^2

Simplifying the equation:

g' = 9.8 * (6,371,000 / 11,471,000)^2
≈ 9.8 * 0.556^2
≈ 9.8 * 0.308

Calculating this expression:

g' ≈ 3.02 m/s^2

Therefore, the effective value of g at 5100 km above the Earth's surface is approximately 3.02 meters/second^2.

Isn't "g" constantly -9.8 m/s/s, regardless of height?

that's what i thought but i put 9.8 and -9.8 in and the answer was wrong. so somehow it's different