State Newton's law of universal gravitation - Leaving Cert Physics - Question 6 - 2008

The force required to keep the ISS in orbit is the gravitational force acting between the Earth and the ISS. This gravitational force provides the necessary centripetal force that keeps the ISS in its circular path.

The direction of this gravitational force is towards the center of the Earth. This inward pull is what keeps the ISS from drifting away into space.

To find the acceleration due to gravity at a height of 400 km above the Earth's surface, we can use the formula: $g' = \frac{G M}{(R + h)^2}$ where:

• $g'$ is the acceleration due to gravity at height $h$,
• $M$ is the mass of the Earth ($6.0 \times 10^{24} \text{kg}$),
• $R$ is the radius of the Earth ($6.4 \times 10^6 ext{m}$), and
• $h = 400 imes 10^3 ext{m}$ (height above Earth’s surface).

Substituting the values gives: $g' = \frac{6.67 \times 10^{-11} \cdot 6.0 \times 10^{24}}{(6.4 \times 10^6 + 400 \times 10^3)^2}$ Calculating this yields the acceleration due to gravity at that height.

An astronaut in the ISS appears weightless because both the ISS and the astronaut are in free fall. The ISS orbits the Earth at a high speed, creating a balance between the gravitational pull of the Earth and the inertia of the ISS's motion. As a result, the astronaut experiences continuous free fall towards Earth, leading to the sensation of weightlessness.

The gravitational force acting on the ISS provides the centripetal force necessary for its circular motion: $F = \frac{m v^2}{r}$ Equating the gravitational force: $G \frac{M m}{r^2} = \frac{m v^2}{r}$ This simplifies to: $v^2 = \frac{G M}{r}$ The period $T$ is related to the speed $v$ and the circumference of the orbit: $T = \frac{2 \pi r}{v}$ Substituting $v$ gives: $T = 2 \pi r \sqrt{\frac{r}{G M}}$ This expresses the relationship between the period $T$, radius $r$, and mass $M$.

Using the derived relationship: $T = 2 \pi \sqrt{\frac{r^3}{G M}}$ Substituting known values:

• $R = 6.4 \times 10^6 ext{m} + 400 \times 10^3 ext{m}$ (total radius above the Earth's surface),
• $G = 6.67 \times 10^{-11} ext{N m}^2 \text{kg}^{-2}$,
• $M = 6.0 \times 10^{24} ext{kg}$. Calculating this will yield the period $T$ for an orbit of the ISS.

The ISS moves in a low Earth orbit, which means as it travels around the Earth, the rotation of the Earth beneath it causes it to pass over different locations on the surface. As the Earth rotates, the ISS orbits at a speed that creates the appearance that it is moving over different points on the Earth throughout its trajectory.

Due to its rapid orbit around the Earth, an astronaut on the ISS experiences approximately 16 sunrises and sunsets each day. This is calculated considering the ISS orbits the Earth roughly every 90 minutes, completing multiple cycles in a 24-hour period.