Define (i) angular velocity, (ii) centripetal force - Leaving Cert Physics - Question 6 - 2005

Centripetal force is the force required to keep an object moving in a circular path, directed towards the center of the circle. It can be described as:

$F_c = \frac{mv^2}{r}$

where $m$ is the mass of the object, $v$ is its tangential velocity, and $r$ is the radius of the circular path.

Newton’s Universal Law of Gravitation states that the gravitational force ($F$) between two masses ($m_1$ and $m_2$) separated by a distance ($r$) is given by:

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

where $G$ is the universal gravitational constant.

For a satellite in circular orbit, the gravitational force provides the necessary centripetal force:

$G \frac{M m}{r^2} = \frac{m v^2}{r}$

where $M$ is the mass of Saturn and $m$ is the mass of the satellite. The velocity $v$ can be expressed as:

$v = \frac{2\pi r}{T}$

Substituting for $v$ gives:

$G \frac{M m}{r^2} = \frac{m (\frac{2\pi r}{T})^2}{r}$

This simplifies to:

$G M = \frac{(2\pi)^2 r}{T^2}$

Rearranging for $T^2$:

$T^2 = \frac{4\pi^2 r^3}{G M}$

This is the desired relationship.

Given that the period $T = 380$ hours, we convert this to seconds:

$T = 380 \times 3600 = 1368000 \, \text{s}$

Using the derived formula:

$T^2 = \frac{4\pi^2 r^3}{G M}$

We rearrange:

$r^3 = \frac{G M T^2}{4\pi^2}$

Substituting values:

• $G = 6.7 \times 10^{-11} \, \text{N m}^2 / \text{kg}^2$
• $M = 5.7 \times 10^{26} \, \text{kg}$
• $T = 1368000$ s

Calculating gives:

$r^3 = \frac{(6.7 \times 10^{-11})(5.7 \times 10^{26})(1368000^2)}{4\pi^2} \approx 1.37 \times 10^{6} \, \text{m}$

Thus:

$r \approx 1.2 \times 10^6 \, \text{m}$.

The time taken ($t$) for the signal to travel a distance ($d$) is given by:

$t = \frac{d}{v}$

where $d = 1.2 \times 10^{12} \, \text{m}$ and $v = 3.0 \times 10^8 \, \text{m/s}$. Thus:

$t = \frac{1.2 \times 10^{12}}{3.0 \times 10^{8}} \approx 4000 \, \text{s} \approx 1.1 \, \text{h}.$

The Doppler effect explains the change in frequency of the received signal. As the satellite moves towards the Earth, the frequency increases (blue shift) because the source is moving closer. Conversely, as it moves away, the frequency decreases (red shift). This is due to the relative motion between the satellite (the source) and the Earth (the detector) affecting the observed frequency.