# Calculate the percentage of Earth’s gravitational force on the ISS astronauts while orbiting at an altitude of 248 miles above Earth’s surface. To do this, compare the gravitational force on an orbiting astronaut to the force on the astronaut here at Earth’s surface using the following equation:

% Fg = ( Fg,orbit / Fg, surface ) x 100 = ( R2 / r2 )x 100

WHERE:

Fg,orbit = force of gravity in orbit from Newton's Law of Gravitation

Fg, surface = force of gravity at Earth’s surface from Newton's Law of Gravitation

R = radius of Earth in km (distance from center of Earth to Earth’s surface at equator)

r = distance from the center of Earth to the center of the object in orbit (R + altitude) in km.

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1. Radius of earth = 6,371 kilometers
248 miles = 399 kilometers
so
r = 6371+399 = 6770

100 (6371/6770)^2 = 88.6 %

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2. By the way, since you set out every other needed step for a solution in your problem statement I assume you do not know how to convert miles to kilometers.
One very crude way is to remember that 30 miles per hour is about 50 kilometers per hour but even easier and more accurate is to type:

248 miles = kilometers

into your Google search box and watch what happens when you hit enter.

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3. According to the question that its asking, the radius of earth is equal to 6,371 kilometers and 248 miles is equal to 399 kilometers, so
r is equal to the equation 6371+399 = 6770, and 100 (6371/6770)^2 = 88.6 %
So if by doing the right amount of math, the Astronauts experience weightlessness in space, while on board the ISS, the experience less tension, and the orbit is at an altitude of 255 miles above Earth’s surface. While at this extravaganza, the gravitational force on an orbiting astronaut to the force on the astronaut here at Earth’s surface is approximately the same velocity of the space flight, where in areas, the ISS orbit is 362 Km, which is in the formula of (3.62 x 124 m) and the earth surface is still at its equinox, and takes it direct radius tension.

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4. yup

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