AT WHAT HEIGHT ABOVE THE EARTH'S SURFACE DOES THE ACCELERATION DUE TO GRAVITY FALL TO 1% OF ITS VALUE AT THE EARTH SURFACE
Solve this equation for height h, in kilometers:
[6378/(6378+h)]^2 = 0.01
Note that an inverse square law is being used.
6378 is the Earth's radius in km.
To find the height above the Earth's surface where the acceleration due to gravity falls to 1% of its value at the Earth's surface, we can use the formula for gravitational acceleration:
\(g' = g \times \left(\frac{R}{R + h}\right)^2\)
Where:
- \(g\) is the acceleration due to gravity at the Earth's surface
- \(g'\) is the acceleration due to gravity at a height above the Earth's surface
- \(R\) is the radius of the Earth
- \(h\) is the height above the Earth's surface
Let's assume the acceleration due to gravity at the Earth's surface \(g\) is approximately 9.8 m/s² and the radius of the Earth \(R\) is approximately 6,371 km. We can plug these values into the formula and solve for \(h\):
\(g' = 0.01g\)
\(0.01g = g \times \left(\frac{R}{R + h}\right)^2\)
Divide both sides by \(g\):
\(0.01 = \left(\frac{R}{R + h}\right)^2\)
Take the square root of both sides:
\(0.1 = \frac{R}{R + h}\)
Multiply both sides by \((R + h)\):
\(0.1(R + h) = R\)
Distribute:
\(0.1R + 0.1h = R\)
Subtract \(0.1R\) from both sides:
\(0.1h = 0.9R\)
Divide both sides by \(0.1\):
\(h = 9R\)
Plugging in the value of \(R\):
\(h = 9 \times 6,371 \text{ km}\)
Calculating this value gives us:
\(h \approx 57,339 \text{ km}\)
Therefore, the height above the Earth's surface at which the acceleration due to gravity falls to 1% of its value at the Earth's surface is approximately 57,339 kilometers.