At what height above the earth's surface. the acceleration due to gravity will be 1/9th of its value at the earth's surface. Re = 6400km
(1/3)^2 = 1/9
so
r = 3 re from earth center=19200
subtract re to get above surface
so 19200-6400= 12800 km
To calculate the height above the Earth's surface where the acceleration due to gravity is 1/9th of its value at the Earth's surface, we can use the inverse square law of gravity.
Let's denote the acceleration due to gravity at the Earth's surface as 'g'. Since g is the acceleration due to gravity at the Earth's surface, we can assume it to be 9.81 m/s^2 (approximately).
Next, we need to calculate the acceleration due to gravity at the desired height, denoted as 'g_h'. According to the inverse square law, the acceleration due to gravity decreases with the square of the distance from the center of the Earth.
So, we can set up the following equation:
g_h = g * (R / (R + h))^2
Where:
g_h is the acceleration due to gravity at height h,
g is the acceleration due to gravity at the Earth's surface,
R is the radius of the Earth (6,400 km or 6,400,000 m),
h is the height above the Earth's surface.
To find the height where g_h is 1/9th of g, we need to solve the equation:
g_h = (1/9) * g
(1/9) * g = g * (R / (R + h))^2
By canceling out the common factors, we get:
1/9 = (R / (R + h))^2
Taking the square root of both sides, we have:
1/3 = R / (R + h)
Now we can solve for h:
(R + h) / R = 3
R + h = 3R
h = 3R - R
h = 2R
Substituting R = 6,400,000 m, we get:
h = 2 * 6,400,000 m
h = 12,800,000 m
Therefore, the height above the Earth's surface where the acceleration due to gravity is 1/9th of its value at the Earth's surface is 12,800,000 meters or 12,800 kilometers.