The distance between the Sun and Mars varies from 2.1 × 10^{11} m to 2.5 × 10^{11} m - AQA - A-Level Physics - Question 13 - 2017 - Paper 2

To determine the force of gravitational attraction when Mars is furthest from the Sun, we can use the formula for gravitational force:

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

In this context, the gravitational force varies inversely with the square of the distance between the two bodies. When Mars is closest to the Sun, let the distance be represented as:

• Closest distance, $r_{min} = 2.1 \times 10^{11} m$.
• Furthest distance, $r_{max} = 2.5 \times 10^{11} m$.

The ratio of the forces can be expressed as:

$\frac{F_{max}}{F_{min}} = \left( \frac{r_{min}}{r_{max}} \right)^2 = \left( \frac{2.1 \times 10^{11}}{2.5 \times 10^{11}} \right)^2$

Calculating this ratio:

$\frac{F_{max}}{F_{min}} = \left( \frac{2.1}{2.5} \right)^2 \approx (0.84)^2 \approx 0.7056$

Thus, the gravitational force when Mars is furthest from the Sun is approximately:

$F_{max} \approx 0.71F$

Therefore, the answer is A. 0.71F.