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

The orbital period $T$ can be determined using the relationship:

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

First, we need to calculate $u$ by substituting $h$ into:

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

Assuming $G = 6.674 \times 10^{-11} \text{N m}^2/\text{kg}^2$, $M = 5.972 \times 10^{24} kg$, and $R = 6.371 \times 10^{6} m$:

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

Calculating $u$:

$\nu = \frac{(6.674 \times 10^{-11})(5.972 \times 10^{24})}{(2.6571 \times 10^{7})^{3/2}}$

Calculating $T$ yields the final answer for the orbital period.

The launch site should ideally be at the equator (site Z), as it allows the satellite to take advantage of the Earth's rotational speed. Launching from the equator provides an initial horizontal velocity, thus requiring less fuel to achieve the necessary orbital speed for the satellite compared to launching from the North Pole (X) or a high mountain (Y), where the effective rotational assist is absent or diminished.

The gravitational potential energy $U$ of the satellite in orbit is given by:

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

Substituting the given values:

• Mass of satellite $m = 1630 kg$
• Use previously calculated $R+h = 2.6571 \times 10^{7} m$;

So,

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

Calculating the values will give the gravitational potential energy.

In a higher orbit, the linear speed of the satellite is smaller than that of a satellite in lower orbit. This is due to the relationship between orbital speed and radius; as the radius increases, the orbital speed decreases since:

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

Thus, for a higher orbit, with larger $R$, the linear speed $v$ will indeed be less than that in the lower orbit described in Question 02.1.