When out in space in International Space Station (ISS), Astronauts experience weightlessness. The ISS’s orbit is 354 km (that is 3.54 x 105 m) above the surface of the earth (You can get more educational NASA resources from: The distance separating the center of the earth from the center of the ISS is then approximately equal to 1.06 times the radius of the earth. What would be in N/kg the magnitude of the gravitational field (g) due to the earth at the position of the space station orbit? Assume 10 N/kg for g at the surface of the earth. Please note that since we are using the 10 N/kg, approximation, we cannot use Newton's universal law formula for solving this question. Compare/Relate the answer to the case of the surface of the earth where g is assume to be 10 N/kg.

Question

When out in space in International Space Station (ISS), Astronauts experience weightlessness. The ISS’s orbit is 354 km (that is 3.54 x 105 m) above the surface of the earth (You can get more educational NASA resources from: The distance separating the center of the earth from the center of the ISS is then approximately equal to 1.06 times the radius of the earth. What would be in N/kg the magnitude of the gravitational field (g) due to the earth at the position of the space station orbit? Assume 10 N/kg for g at the surface of the earth. Please note that since we are using the 10 N/kg, approximation, we cannot use Newton's universal law formula for solving this question. Compare/Relate the answer to the case of the surface of the earth where g is assume to be 10 N/kg.

Answer 1

To find the magnitude of the gravitational field (g) due to the Earth at the position of the International Space Station (ISS) orbit, we can use the concept of gravitational acceleration.

First, let's set up the equation using the given information:
- g(surface) = 10 N/kg (gravitational field at the surface of the Earth)
- R(earth) = radius of the Earth

The distance separating the center of the Earth from the center of the ISS is given as approximately 1.06 times the radius of the Earth. So, the distance can be calculated as:
d(ISS) = 1.06 * R(earth)

Now, let's find the magnitude of the gravitational field at the position of the ISS orbit.

We know that the gravitational field strength depends on the distance from the center of the Earth. The formula for gravitational acceleration at a distance r from the center of the Earth can be given as:

g(r) = (G * M(earth)) / r^2

where
- G is the gravitational constant (approximately 6.67430 x 10^-11 N m^2/kg^2)
- M(earth) is the mass of the Earth (approximately 5.972 x 10^24 kg)
- r is the distance from the center of the Earth

Since the ISS is in the orbit around the Earth, the distance r is equal to the distance between the center of the Earth and the ISS orbit. So, we can substitute this value into the above formula:

g(ISS) = (G * M(earth)) / d(ISS)^2

Substituting the given values and calculating:

g(ISS) = (6.67430 x 10^-11 N m^2/kg^2 * 5.972 x 10^24 kg) / (1.06 * R(earth))^2

Now, let's compare/relate the answer to the case of the surface of the Earth where g is assumed to be 10 N/kg.

The magnitude of the gravitational field at the ISS orbit can be found using the above equation, and it will be different from the value of 10 N/kg at the surface of the Earth.

By substituting the values, the magnitude of the gravitational field at the ISS orbit can be calculated in N/kg. The answer will give you the actual value of the gravitational field strength experienced by the astronauts in the ISS orbit relative to the value at the Earth's surface.