The Global Positioning System (GPS) uses satellites to support navigation on Earth - AQA - A-Level Physics - Question 2 - 2021 - Paper 2

To calculate the orbital period $T$, we use:

$T = \frac{2\pi}{\omega}$

Substituting our expression for $\omega$:

$T = \frac{2\pi (R+h)^{3/2}}{\sqrt{GM}}$

Plugging in the values:

• $R = 6.371 \times 10^{6} \, m$
• $h = 2.02 \times 10^{7} \, m$
• $G = 6.674 \times 10^{-11} \, m^{3}kg^{-1}s^{-2}$
• $M = 5.972 \times 10^{24} \, kg$

Calculating:

$h = 2.02 \times 10^{7} + 6.371 \times 10^{6} = 2.6571 \times 10^{7} \, m$

Then substituting values to compute $T$.

The optimal launch site is $Z$, which is on the equator. Launching from the equator allows the satellite to take advantage of the Earth's rotational speed, reducing the amount of fuel needed to achieve orbit. Launching from higher latitudes, like $X$ at the North Pole, requires additional fuel to achieve the necessary speed and trajectory.

The gravitational potential energy $U$ is given by:

$U = -\frac{GMm}{R+h}$

Using the values:

• $G = 6.674 \times 10^{-11} \, m^{3}kg^{-1}s^{-2}$
• $M = 5.972 \times 10^{24} \, kg$
• $m = 1630 \, kg$
• $R + h = 2.6571 \times 10^{7} \, m$

Substituting these into the equation gives:

$U = -\frac{6.674 \times 10^{-11} \cdot 5.972 \times 10^{24} \cdot 1630}{2.6571 \times 10^{7}}$

Now compute value for $U$.

In a higher orbit, the linear speed of the satellite is smaller. The orbital speed $v$ depends on the radius of the orbit as given by:

$v = \sqrt{\frac{GM}{R+h}}$

As $R+h$ increases for a satellite in a higher orbit, the linear speed decreases, leading to a slower orbital motion compared to the satellite in Question 02.1.