State Newton's law of universal gravitation - Leaving Cert Physics - Question 6 - 2013

Angular velocity ($\omega$) is defined as the rate of change of the angle of an object moving along a circular path. It can be expressed mathematically as:

$\omega = \frac{\theta}{t}$

where $\theta$ is the angle in radians and $t$ is the time.

When an object moves in a circular path, the relationship between angular velocity ($\omega$) and linear velocity ($v$) is given by:

$v = r \omega$

where $r$ is the radius of the circular path. Rearranging this gives:

$\omega = \frac{v}{r}$

To find the angular velocity ($\omega$) of the ISS, we first need to calculate the period of the ISS's orbit. The given altitude is 4.13 x 10^6 m, and the orbital time is 92 minutes 50 seconds, which we convert into seconds:

$t = 92 \text{ minutes} + 50 \text{ seconds} = 92 \times 60 + 50 = 5550 \text{ seconds}$

Now we can calculate the angular velocity using:

$\omega = \frac{2\pi}{t} = \frac{2\pi}{5550} \approx 1.131 \times 10^{-3} \text{ rad/s}$

Next, the linear velocity ($v$) can be calculated as:

$v = \frac{2\pi \times (4.13 \times 10^6)}{t} \approx 761.5 \text{ m/s}$

The ISS experiences centripetal (or gravitational) acceleration as it travels in a circular orbit around the Earth. This acceleration is provided by the gravitational force between the Earth and the ISS.

The attractive force ($F$) between the Earth and the ISS can be calculated using Newton's law of gravitation:

$F = \frac{G m_1 m_2}{r^2}$

where $m_1$ is the mass of the Earth, $m_2$ is the mass of the ISS, and $r$ is the distance from the center of the Earth to the ISS.

Using the known values:

• Mass of ISS $m_2 = 4.5 \times 10^5 \, \text{kg}$
• Gravitational constant $G = 6.674 \times 10^{-11} \, \text{m}^3/\text{kg.s}^2$
• Radius of the Earth $R = 6.37 \times 10^6 \, \text{m}$
• Distance $r = R + 4.13 \times 10^6 = 10.5 \times 10^6 \text{ m}$

The attractive force:

$F = \frac{(6.674 \times 10^{-11}) \times (5.95 \times 10^{24}) \times (4.5 \times 10^5)}{(10.5 \times 10^6)^2} \approx 3.884 \times 10^6 \text{ N}$

To find the mass of the Earth:

$M = \frac{F r^2}{G m_2}$

Substituting the known values allows for calculating the mass.

The occupants of the ISS experience apparent weightlessness because both they and the ISS are in free fall towards the Earth. Although the gravitational acceleration on the ISS is 8.63 m/s², the ISS is continuously falling towards Earth while also moving forward, creating a state of microgravity for the inhabitants.

A geostationary communications satellite orbits the Earth at a height where its period is the same as the Earth's rotational period, which is approximately 24 hours. Therefore, the period of a geostationary communications satellite is:

1 day.