Earths mass is 6 x 10^24kg and its radius is 6.38 x 10^6m
(6380 km ). Use the inverse-square law to show that in space-shuttle territory, 200 km above the earths surface the force of gravity on a shuttle is about 94% of that at the earths surface.
Let the acceleration of gravity at 200 km altitude be g' and the value at sea level be g. Then
g'/g = (6380/6580)^2 = 0.94013
The rounds off to 94%, to the nearest 1%.
To demonstrate the inverse-square law and calculate the force of gravity at a given distance from the Earth's surface, we can use the formula:
F = (G * M * m) / r^2
Where:
F is the force of gravity
G is the universal gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^−1 s^−2)
M is the mass of the Earth (6 x 10^24 kg)
m is the mass of the object experiencing the gravitational force (assumed to be the space shuttle)
r is the distance between the center of the Earth and the object (radius + altitude)
Given that the mass of the Earth (M) is 6 x 10^24 kg and the radius of the Earth (r) is 6.38 x 10^6 m (6380 km), we need to calculate the force of gravity at the Earth's surface and at the given distance (200 km above the surface). Let's first calculate the force at the Earth's surface.
F_surface = (G * M * m) / r_surface^2
Where r_surface is the radius of the Earth.
F_surface = (6.67430 × 10^-11 * 6 x 10^24 * m) / (6.38 x 10^6)^2
Now, let's calculate the force at the given distance above the Earth's surface.
F_altitude = (G * M * m) / r_altitude^2
Where r_altitude = r_surface + altitude
F_altitude = (6.67430 × 10^-11 * 6 x 10^24 * m) / (r_surface + 2 x 10^5)^2
To find the ratio of the force at the altitude to the force at the Earth's surface:
Ratio = F_altitude / F_surface
Ratio = [(6.67430 × 10^-11 * 6 x 10^24 * m) / (r_surface + 2 x 10^5)^2] / [(6.67430 × 10^-11 * 6 x 10^24 * m) / (6.38 x 10^6)^2]
Simplifying the equation:
Ratio = [(r_surface + 2 x 10^5)^2] / [(6.38 x 10^6)^2]
Plug in the given values and calculate the ratio:
Ratio = [(6.38 x 10^6 + 2 x 10^5)^2] / [(6.38 x 10^6)^2]
After calculating the above expression, we find that the ratio of the force of gravity at an altitude of 200 km above the Earth's surface to that at the Earth's surface is approximately 0.94 (or 94%). Therefore, the force of gravity on a space shuttle in that territory is about 94% of the force at the Earth's surface.