Figure 2 shows a moon of mass m in a circular orbit of radius r around a planet of mass M, where m << M - AQA - A-Level Physics - Question 2 - 2020 - Paper 2

Using the data from Table 2, we know:

Miranda's orbital radius is given by:

$T^2 = k r^3,$ with T for Miranda being 1.41 days, which converted to seconds gives:

$T = 1.41 \times 24 \times 3600 \approx 121,464 \text{ seconds}$.

Calculating k using Miranda's data:

$k = \frac{121464^2}{(1.29 \times 10^8)^3}.$

Substituting k into the equation for Umbriel:

$4.14^2 = k X^3,$ we can solve for X. By calculating:

$X = \sqrt[3]{\frac{4.14^2}{k}},$ and you will arrive at the result for X.

Once we have X for Umbriel, we can substitute this value back into the equation:

Using the formula for k calculated previously:

$k = \frac{4\pi^2}{GM}.$

Rearranging this gives:

$M = \frac{4\pi^2}{k}.$

Substituting the previously calculated k value yields the mass of Uranus:

$M = \frac{4\pi^2}{k}.$ Calculate to find M.

The escape velocity v_e for an object on the surface of a moon is given by:

$v_e = \sqrt{\frac{2GM}{R}},$

where R is the radius of the moon. Each mass and diameter can be used to derive R:

• For Ariel, using mass and diameter, we find R.
• Repeat for Oberon and Titania.

Comparing the escape velocities calculated using the above formula will give the moon with the highest escape velocity.

Using gravitational potential energy:

$E_p = mgh,$ where m is mass, g is acceleration due to gravity, and h is height.

The mechanism needs to provide sufficient energy for heights greater than 100 m. The kinetic energy at the surface must be sufficient to convert into potential energy:

$E_k = mgh_{max} > 100m imes g_a,$ where g_a is the gravity on Ariel.

By optimizing the spring mechanism, ensuring it compresses sufficiently to provide energy greater than the gravitational potential energy required for 100 m height will achieve this.