6. (i) State Newton's law of universal gravitation - Leaving Cert Physics - Question 6 - 2013

Angular velocity ($\omega$) is a measure of the rate of change of the angle of an object rotating around a central point. It is defined as the angle ($\theta$) swept per unit time ($t$):

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

For an object moving in a circle, the relationship between angular velocity and linear velocity ($v$) is given by:

$v = r \omega$

where $r$ is the radius of the circle. Rearranging this equation gives:

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

(a) To find the angular velocity, we first need to convert the orbital period of the ISS from minutes to seconds:

$T = 92 \text{ mins} \times 60 \text{ s/min} = 5520 \text{ s}$

Now, using the formula for angular velocity:

$\omega = \frac{2\pi}{T} = \frac{2\pi}{5520} \approx 1.14 \times 10^{-3} \text{ rad/s}$

(b) The linear velocity is calculated using the relationship:

$v = r \omega$

Using $r = 4.13 \times 10^7 \text{ m}$,

$v = (4.13 \times 10^7)(1.14 \times 10^{-3}) \approx 47176.83 \text{ m/s}$

The ISS experiences centripetal acceleration as it travels in a circular orbit around the Earth. This acceleration is provided by the gravitational force exerted by the Earth on the ISS.

To calculate the attractive force, we use Newton's law of universal gravitation:

$F = G \frac{m_{ISS} m_E}{r^2}$

where:

• $G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2$
• $m_{ISS} = 4.5 \times 10^6 \text{ kg}$
• $r=4.13 \times 10^7 + 6.37 \times 10^6 = 4.76 \times 10^7 \text{ m}$

Substituting the values:

$F \approx 3.84 \times 10^6 \text{ N}$

Hence, to find the mass of the earth ($M$):

$M = \frac{F r^2}{G m_{ISS}} = \frac{(3.84 \times 10^6)(4.76 \times 10^7)^2}{6.674 \times 10^{-11} (4.5 \times 10^6)} \approx 5.95 \times 10^{24} \text{ kg}$

Occupants of the ISS experience apparent weightlessness because they and the station are in free fall towards the Earth. Both the ISS and its occupants accelerate at the same rate due to gravity (8.63 m/s²), creating a sensation of weightlessness as there is no normal force acting on them.

The period ($T$) of a geostationary satellite is 24 hours or 1 day. This means that it takes the satellite 24 hours to complete one full orbit around the Earth, matching the rotation period of the Earth itself.