In the circular orbit of a satellite around the Earth, the required centripetal force is the gravitational force between the satellite and the Earth - Leaving Cert Physics - Question 6 - 2015

Newton's law of universal gravitation states that the force of gravity between two masses is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, this is expressed as:

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

where $F$ is the gravitational force, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses, and $r$ is the distance between the centers of the two masses.

To derive this relationship, we start from the equation for centripetal force and the gravitational force:

1. Centripetal force: $F_c = \frac{m v^2}{r}$
2. Gravitational force: $F_g = G \frac{m M}{r^2}$

Where:

• $m$ is the mass of the satellite,
• $M$ is the mass of the Earth,
• $v = \frac{2 \pi r}{T}$, where $T$ is the orbital period.

Equating centripetal force to gravitational force:

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

Cancelling $m$ and rearranging gives:

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

Substituting for velocity:

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

This leads to:

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

From which we can express the period:

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

Given:

• The orbital radius of a GPS satellite is $r = 2.66 \times 10^7 \text{ m}$
• The radius of the Earth is $R_E = 6.371 \times 10^6 \text{ m}$

The height $h$ above the Earth's surface can be calculated as:

$h = r - R_E = 2.66 \times 10^7 - 6.371 \times 10^6 \approx 2.023 \times 10^7 \text{ m}$

Using the relation for orbital speed:

$v = \sqrt{\frac{GM}{r}}$

Substituting $G \approx 6.674 \times 10^{-11} \text{m}^3 \text{kg}^{-1} \text{s}^{-2}$ and $M = 5.97 \times 10^{24} \text{kg}$, we find:

$v \approx \sqrt{\frac{(6.674 \times 10^{-11})(5.97 \times 10^{24})}{2.66 \times 10^7}} \approx 3.869 \times 10^7 \text{ m/s}$

Using the formula for time:

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

Where:

• $d$ is the distance (which is the height of the satellite above the Earth), approximately equal to $h = 2.023 \times 10^7$ m,
• and $v = 3.869 \times 10^7$ m/s:

Substituting values gives:

$t = \frac{2.023 \times 10^7}{3.869 \times 10^7} \approx 0.067 \text{ s}$

GPS satellites are not classed as geostationary satellites because they do not maintain a fixed position relative to the Earth's surface. Instead, they orbit the Earth every 12 hours, allowing them to cover the entire surface over time. Geostationary satellites, in contrast, have an orbital period equal to the Earth's rotation period, which is 24 hours.

The type of electromagnetic radiation that has the next lowest frequency after radio waves is microwaves.