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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

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Figure 2 shows a moon of mass m in a circular orbit of radius r around a planet of mass M, where m << M. The moon has an orbital period T. T is related to r by $$T... show full transcript

Worked Solution & Example Answer: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

Step 1

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

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Answer

To demonstrate this relationship, we start from the equation for the orbital period:

T2=kr3T^2 = k r^3

Rearranging for k gives:

k=T2r3k = \frac{T^2}{r^3}.

Next, consider the gravitational force acting on the moon:

F=GMmr2 and F=mv2r, where v=2πrTF = \frac{GMm}{r^2} \text{ and } F = \frac{mv^2}{r}, \text{ where } v = \frac{2\pi r}{T}.

Setting these two forces equal yields:

GMmr2=m(2Ï€r/T)2r\frac{GMm}{r^2} = \frac{m(2\pi r/T)^2}{r}.

Simplifying leads to:

GM=4Ï€2rT2.GM = \frac{4\pi^2r}{T^2}.

Substituting this back into our expression for k gives:

k=4Ï€2GM.k = \frac{4\pi^2}{GM}.

Step 2

Calculate the orbital radius X of Umbriel.

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Answer

Using the data from Table 2, we know:

Miranda's orbital radius is given by:

T2=kr3,T^2 = k r^3, with T for Miranda being 1.41 days, which converted to seconds gives:

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

Calculating k using Miranda's data:

k=1214642(1.29×108)3.k = \frac{121464^2}{(1.29 \times 10^8)^3}.

Substituting k into the equation for Umbriel:

4.142=kX3,4.14^2 = k X^3, we can solve for X. By calculating:

X=4.142k3,X = \sqrt[3]{\frac{4.14^2}{k}}, and you will arrive at the result for X.

Step 3

Calculate the mass of Uranus.

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Answer

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

Using the formula for k calculated previously:

k=4Ï€2GM.k = \frac{4\pi^2}{GM}.

Rearranging this gives:

M=4Ï€2k.M = \frac{4\pi^2}{k}.

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

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

Step 4

Deduce which moon in Table 3 has the greatest escape velocity for an object on its surface.

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Answer

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

ve=2GMR,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.

Step 5

Determine how the mechanism could project the object vertically to a maximum height greater than 100 m when placed on the surface of Ariel.

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Answer

Using gravitational potential energy:

Ep=mgh,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:

Ek=mghmax>100mimesga,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.

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