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.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 Newtons the magnitude of the gravitational force that the earth exerts on a 70 kg Astronaut when in the space station?
To calculate the magnitude of the gravitational force that the Earth exerts on a 70 kg astronaut in the space station, we can use the universal law of gravitation.
The formula for the magnitude of the gravitational force is:
F = (G * m1 * m2) / r^2
Where:
F is the gravitational force,
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, the Earth and the astronaut),
and r is the distance between the centers of the two objects.
Given:
Mass of the Earth (M) = 5.972 × 10^24 kg,
Mass of the astronaut (m) = 70 kg,
Distance between the centers of the Earth and the ISS (r) = 1.06 times the radius of the Earth.
First, we need to find the radius of the Earth, which is approximately 6.371 × 10^6 m.
Now, we can calculate the distance between the centers of the Earth and the ISS:
r = 1.06 * (radius of the Earth) = 1.06 * 6.371 × 10^6 m = 6.75326 × 10^6 m.
Plugging the values into the formula, we have:
F = (6.67430 × 10^-11 N m^2/kg^2 * (5.972 × 10^24 kg) * (70 kg)) / (6.75326 × 10^6 m)^2.
Simplifying the equation, we get:
F ≈ 6.3056 × 10^2 N.
Therefore, the magnitude of the gravitational force that the Earth exerts on a 70 kg astronaut in the space station is approximately 630.56 Newtons.