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? Use 10 N/kg for g at the surface of the earth.

To calculate the magnitude of the gravitational force that the earth exerts on the astronaut in the ISS, we can use the formula for gravitational force:

F = (G * m1 * m2) / r^2

where F is the gravitational force, G is the gravitational constant (approximately 6.67 x 10^-11 N·(m/kg)^2), m1 and m2 are the masses of the two objects (in this case, the mass of the astronaut and the mass of the Earth), and r is the distance between the centers of the two objects.

In this case, the mass of the astronaut is 70 kg.

The mass of the Earth can be considered to be concentrated at its center, so the distance between the center of the Earth and the center of the ISS is given as 1.06 times the radius of the Earth. The radius of the Earth is approximately 6.37 x 10^6 m.

Plugging these values into the equation, we get:

F = (6.67 x 10^-11 N·(m/kg)^2 * 70 kg * 5.97 x 10^24 kg) / (1.06 * 6.37 x 10^6 m)^2

Calculating this equation will give us the magnitude of the gravitational force that the Earth exerts on the astronaut in Newtons.