Many people mistakenly believe that astronauts who orbit the Earth are "above gravity." The mass of the Earth is 6x10 ^ 34 kg and radius is 6.38x10 ^ 6 m (6380 km) Use Newton's law to show that in the territory of the special ferry, 200 km above the earth's surface, the force of gravity on the shuttle is approximately 94% that of the Earth's surface
gravity follows an inverse-square relation
[6380 / (6380 + 200)]^2 = .940
To calculate the force of gravity on the shuttle, we need to use Newton's law of universal gravitation. This law states that the force of gravity between two objects is directly proportional to their masses and inversely proportional to the square of the distance between their centers.
The formula for Newton's law of universal gravitation is:
F = (G * m1 * m2) / r^2
where F is the force of gravity, G is the gravitational constant (6.67 × 10^-11 N m^2/kg^2), m1 and m2 are the masses of the two objects, and r is the distance between their centers.
In this case, we are interested in finding the force of gravity on the shuttle, which is 200 km (or 200,000 m) above the Earth's surface. So, we need to calculate the force using the following values:
Mass of the Earth (m1) = 6 × 10^24 kg
Mass of the shuttle (m2) = will be neglected since it is very small compared to the Earth's mass.
Distance from the Earth's center to the shuttle (r) = radius of the Earth (6.38 × 10^6 m) plus the height of the shuttle (200,000 m)
Plugging these values into the formula, we get:
F = (G * m1 * m2) / r^2
= (G * m1 * 1) / r^2 (neglecting the mass of the shuttle)
= (6.67 × 10^-11 N m^2/kg^2 * 6 × 10^24 kg) / (6.58 × 10^6 m)^2
≈ 9.45 × 10^13 N
Now, we need to compare this force to the force of gravity on the Earth's surface.
The force of gravity on the Earth's surface can be calculated using the same formula with the following values:
Mass of the Earth (m1) = 6 × 10^24 kg
Distance from the Earth's center to the surface (r) = radius of the Earth (6.38 × 10^6 m)
Plugging these values into the formula, we get:
F_surface = (G * m1 * 1) / r^2
= (6.67 × 10^-11 N m^2/kg^2 * 6 × 10^24 kg) / (6.38 × 10^6 m)^2
≈ 9.82 × 10^13 N
Now, we can calculate the percentage of the force on the shuttle compared to the Earth's surface:
Percentage = (F / F_surface) * 100
= (9.45 × 10^13 N / 9.82 × 10^13 N) * 100
≈ 96.15 %
Therefore, the force of gravity on the shuttle 200 km above the Earth's surface is approximately 96.15% of the force of gravity on the Earth's surface.
To show that the force of gravity on a shuttle 200 km above the Earth's surface is approximately 94% of the force on the Earth's surface, we can use Newton's law of universal gravitation.
Newton's law states that the gravitational force between two objects is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, the formula can be written as:
F = G * (m1 * m2) / r^2
Where:
F is the gravitational force between the two objects,
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2),
m1 and m2 are the masses of the two objects, and
r is the distance between the centers of the two objects.
Now, let's calculate the force of gravity on the shuttle 200 km above the Earth's surface. We'll compare it to the force of gravity on the Earth's surface.
The mass of the Earth (m1) is 6x10^24 kg, and the mass of the shuttle doesn't affect the force significantly, so we can ignore it for this calculation. The radius of the Earth (r1) is 6.38x10^6 m.
First, we need to calculate the radius when the shuttle is 200 km above the Earth's surface. Adding the distance of 200 km to the Earth's radius gives us the new radius:
r2 = r1 + altitude
= 6.38x10^6 m + 200 km
= 6.38x10^6 m + 200,000 m
= 6.58x10^6 m
Now, we can calculate the gravitational force on the shuttle using the modified radius:
F2 = G * (m1 * m2) / r2^2
We can compare this force to the gravitational force on the Earth's surface (F1). The force at the Earth's surface is given by:
F1 = G * (m1 * m2) / r1^2
To find the percentage of F2 in relation to F1, we divide F2 by F1:
(F2 / F1) * 100
Substituting the values:
(F2 / F1) * 100 = [G * (m1 * m2) / r2^2] / [G * (m1 * m2) / r1^2]
= (r1^2 / r2^2) * 100
Plugging in the values of r1 = 6.38x10^6 m and r2 = 6.58x10^6 m:
(F2 / F1) * 100 = (6.38x10^6 m)^2 / (6.58x10^6 m)^2 * 100
≈ 0.9404 * 100
≈ 94.04%
Hence, the force of gravity on the shuttle 200 km above the Earth's surface is approximately 94% of the force at the Earth's surface.