Compared with the strength of Earth's gravity at its surface, how much weaker is gravity at a distance of 13 Earth radii from Earth's center?
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To determine the strength of gravity at a specific distance from Earth's center, we can use Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
Where:
F is the force of gravity
G is the gravitational constant (approximately 6.67430 × 10^-11 N m^2 / kg^2)
m1 and m2 are the masses of the two objects (in this case, Earth and an object at the given distance)
r is the distance between the centers of the two objects
In this case, we are comparing the strength of gravity at two different distances. Let's assume that the mass of the object remains the same.
At the Earth's surface, the distance from the center of the Earth is the radius of the Earth, which is approximately 6,371 kilometers (or 6,371,000 meters).
At a distance of 13 Earth radii from the center of the Earth, the distance from the center of the Earth would be 13 times the radius of Earth, which is 13 * 6,371,000 meters = 82,723,000 meters.
Now, we can calculate the ratio of the strengths of gravity at these two distances:
F1 / F2 = (G * m1 * m2 / r1^2) / (G * m1 * m2 / r2^2)
The masses, m1 and m2, cancel out, leaving us with:
F1 / F2 = r2^2 / r1^2
Plugging in the values:
F1 / F2 = (82,723,000)^2 / (6,371,000)^2
Calculating this ratio, we find that the strength of gravity at a distance of 13 Earth radii from Earth's center is approximately 0.0203 times weaker compared to the gravity at Earth's surface.
Therefore, the gravity at a distance of 13 Earth radii from Earth's center is approximately 1/49th (or 1/0.0203) of the strength of gravity at Earth's surface.