Question

The International Space Station (ISS) orbits Earth in a nearly circular orbit that is 345 km above Earth's surface.

1)How many hours does it take for the ISS to make each orbit? (Express your answer to three significant figures.)

2)Some of the experiments performed by astronauts in the ISS involve the effects of "weightlessness" on objects. What gravitational force does Earth exert on a 13.0-kg object in the ISS? Express your answer in newtons. (Express your answer to three significant figures.)

3)What gravitational force does Earth exert on a 13.0-kg object in the ISS? Express your answer as a fraction of the force that Earth would exert on the object at Earth's surface. (Express your answer to three significant figures.) (in %)

Answer 1

The International Space Station orbits Earth every 90 minutes. The gravitational force on a 13.0-kg object in the ISS is approximately 114.66 N, which is 88.0% of the force that would be experienced at Earth's surface.

Step-by-step explanation:

The student has asked about the orbital period of the International Space Station (ISS), the gravitational force on an object in the ISS, and how that force compares to the force on Earth’s surface.

Orbital Period of the ISS

The ISS completes each orbit around Earth in approximately 90 minutes. This orbital period is consistent across various sources including NASA and photographic evidence from space missions.

Gravitational Force on a 13.0-kg Object Aboard the ISS

To approximate the force of gravity on a 13.0-kg object aboard the ISS, we consider that the gravitational force is about 90 percent of what it is at Earth's surface. Given that the acceleration due to gravity at the surface is approximately 9.80 m/s², the gravitational force at the ISS's altitude is around 8.82 m/s² (90% of 9.80 m/s²). Therefore, the force on a 13.0-kg object would be:

Force = mass × acceleration = 13.0 kg × 8.82 m/s² = 114.66 N (Newton)

Gravitational Force as a Fraction of Earth's Surface Force

The gravitational force on a 13.0-kg object in the ISS as a fraction of the force at Earth's surface, to three significant figures, would be 88.0%. This can be calculated as:

Fraction = (force on object in ISS) / (force on object at Earth's surface) = (13.0 kg × 8.82 m/s²) / (13.0 kg × 9.80 m/s²) = 0.880, or 88.0%