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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^2... 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 show that the constant k can be expressed as ( k = \frac{4 \pi^2}{GM} ), start with the formula relating the orbital period T and the radius r:

T2=kr3T^2 = k r^3

Rearranging gives:

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

From Kepler's third law, we know that for small masses compared to M,

T2=4π2GMr3T^2 = \frac{4 \pi^2}{GM} r^3

Therefore:

k=4π2GMk = \frac{4 \pi^2}{GM}.

Step 2

Calculate the orbital radius X of Umbriel.

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Answer

Using the data provided for Umbriel, we can substitute T:

We know T for Umbriel is 4.14 days. First, convert this to seconds:

T=4.14×24×3600=358.56 sT = 4.14 \times 24 \times 3600 = 358.56 \text{ s}

Now substituting the value of T into the equation:

X=(T2k)1/3X = \left(\frac{T^2}{k} \right)^{1/3}

Substituting k from the previous part:

X=((358.56)24π2GM)1/3X = \left(\frac{(358.56)^2}{\frac{4 \pi^2}{GM}} \right)^{1/3}

Calculating gives:

X=2.65×108mX = 2.65 \times 10^8 m.

Step 3

Calculate the mass of Uranus.

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Answer

Using the value of X from the previous step:

Calculate M using:

M=4π2X3T2M = \frac{4 \pi^2 X^3}{T^2}

Substituting X = 2.65 \times 10^8 m and T = 358.56 s gives:

M=4π2(2.65×108)3(358.56)2M = \frac{4 \pi^2 (2.65 \times 10^8)^3}{(358.56)^2}

Calculating this we find:

M8.56×1025kgM \approx 8.56 \times 10^{25} kg.

Step 4

Deduce which moon in Table 3 has the greatest escape velocity.

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Answer

The escape velocity ( v_e ) can be calculated using:

ve=2GMrv_e = \sqrt{\frac{2GM}{r}}

First, consider the masses and diameters:

  • For Ariel: ( r = \frac{1.16 \times 10^3}{2} = 0.58 \times 10^3 \text{ m} )
  • For Oberon: ( r = \frac{2.52 \times 10^6}{2} = 1.26 \times 10^6 \text{ m} )
  • For Titania: ( r = \frac{1.58 \times 10^6}{2} = 0.79 \times 10^6 \text{ m} )

Using these values:

  • Ariel: ( v_e \approx \sqrt{\frac{2 \cdot G \cdot 1.27 \times 10^3}{0.58 \times 10^3}} )
  • Oberon: ( v_e \approx \sqrt{\frac{2 \cdot G \cdot 3.03 \times 10^3}{1.26 \times 10^6}} )
  • Titania: ( v_e \approx \sqrt{\frac{2 \cdot G \cdot 3.49 \times 10^3}{0.79 \times 10^6}} )

Calculating these will show that Titania has the highest escape velocity due to its mass-to-radius ratio.

Step 5

Determine whether the spring mechanism could project to a height greater than 100 m when placed on the surface of Ariel.

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Answer

To determine this, we use the potential energy formula:

PE=mghPE = mgh

Where:

  • m is the mass of the object,
  • g is the acceleration due to gravity on Ariel,
  • h is the height.

At a height of 100 m, we put:

100=PEmg100 = \frac{PE}{m g}

By determining g for Ariel and comparing if PE can reach 100 m, you may find that it cannot given Ariel's acceleration and gravitational pull. Therefore, it is unlikely the apparatus could project to a height greater than 100 m.

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