What is the gravity at the height of 15000 km above the earth surface
The Earth's radius is Re = 6357 km. A location 15,000 km above the surface would be
Re + 15,000 = 21,357 km
from the center.
Since the accleration of gravity, g, is proportional to 1/R^2, where R is the distance from the center of the earth, at that height
g = (6357/21,357)^2 * 9.8 m/s^2
= 0.87 m/s^2
To calculate the gravity at a certain height above the Earth's surface, we can use the formula for the acceleration due to gravity:
g = G * M / r^2
Where:
g is the acceleration due to gravity,
G is the gravitational constant (approximately 6.674 × 10^-11 N(m/kg)^2),
M is the mass of the Earth (approximately 5.972 × 10^24 kg),
and r is the distance from the center of the Earth.
In this case, the height is given as 15000 km above the Earth's surface. We need to convert this height to the distance from the center of the Earth by adding the Earth's radius (approximately 6371 km) to the given height:
r = 15000 km + 6371 km = 21371 km
Now, we need to convert this distance from kilometers to meters:
r = 21371 km * 1000 m/km = 21371000 m
Plugging these values into the formula, we can calculate the gravity at a height of 15000 km above the Earth's surface:
g = (6.674 × 10^-11 N(m/kg)^2 * 5.972 × 10^24 kg) / (21371000 m)^2
Calculating this, we find:
g ≈ 1.058 m/s^2
Therefore, the gravity at a height of 15000 km above the Earth's surface is approximately 1.058 m/s^2.