A communications satellite orbits the Earth at a height of 36.0 × 10^6 m above the surface of the Earth - Scottish Highers Physics - Question 4 - 2019

We use Newton's law of gravitation which states that the gravitational force between two masses is given by:

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

Where:

• $F$ is the gravitational force (57 N)
• $G$ is the gravitational constant, approximately $6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$
• $m_1$ is the mass of the Earth ($6.0 \times 10^{24} \text{ kg}$)
• $m_2$ is the mass of the satellite (unknown)
• $r$ is the distance from the centre of the Earth to the satellite ($42.4 \times 10^6 \text{ m}$)

Plugging the known values into the equation, we solve for $m_2$:

$57 = 6.67 \times 10^{-11} \frac{(6.0 \times 10^{24}) m_2}{(42.4 \times 10^6)^2}$

Rearranging and substituting to find $m_2$ gives:

$m_2 = \frac{57 (42.4 \times 10^6)^2}{6.67 \times 10^{-11} (6.0 \times 10^{24})} \approx 260 ext{ kg}$

The gravitational field strength $g$ at the satellite can be determined using the formula:

$g = \frac{F}{m_s}$

Where:

• $F$ is the gravitational force (57 N)
• $m_s$ is the mass of the satellite (calculated to be 260 kg)

Therefore:

$g = \frac{57 ext{ N}}{260 ext{ kg}} \approx 0.22 ext{ N/kg}$

The gravitational force of attraction between the second satellite and the Earth is lower than that of the first satellite because gravitational force depends on mass. Since the second satellite has a quarter of the mass of the first satellite, the force can be expressed as:

$F_2 = \frac{1}{4} F_1$

This shows that the gravitational force of attraction is reduced by the same factor, thereby justifying that if the first satellite's force is $F_1$, the second satellite's force of attraction will be $\frac{1}{4}F_1$.