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A planet has radius $R$ and density $\rho$ - AQA - A-Level Physics - Question 12 - 2021 - Paper 2

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A planet has radius $R$ and density $\rho$. The gravitational field strength at the surface is $g$. What is the gravitational field strength at the surface of a pl... show full transcript

Worked Solution & Example Answer:A planet has radius $R$ and density $\rho$ - AQA - A-Level Physics - Question 12 - 2021 - Paper 2

Step 1

What is the gravitational field strength at the surface of a planet of radius $2R$ and density $2\rho$?

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Answer

To find the gravitational field strength at the surface of the planet with new radius and density, we can use the formula for gravitational field strength:

g′=GMR2g' = \frac{GM}{R^2}

  1. Calculate the new mass:

    • The volume of the planet is given by:
      V=43πR3V = \frac{4}{3}\pi R^3
    • For the new planet with radius 2R2R:
      V′=43π(2R)3=43π(8R3)=8⋅43πR3V' = \frac{4}{3}\pi (2R)^3 = \frac{4}{3}\pi (8R^3) = 8 \cdot \frac{4}{3}\pi R^3
    • The new mass M′M' with density 2ρ2\rho:
      M′=V′⋅2ρ=8⋅43πR3imes2ρ=643πR3⋅ρM' = V' \cdot 2\rho = 8 \cdot \frac{4}{3}\pi R^3 imes 2\rho = \frac{64}{3}\pi R^3 \cdot \rho

  2. Calculate the new gravitational field strength:

    • Substituting for M′M' and RR into the gravitational field strength formula:
      g′=G⋅643πR3⋅ρ(2R)2g' = \frac{G \cdot \frac{64}{3}\pi R^3 \cdot \rho}{(2R)^2}
    • Simplifying:
      g′=643⋅G⋅π⋅ρR34R2=163⋅G⋅π⋅ρR3R2=163gg' = \frac{64}{3} \cdot \frac{G \cdot \pi \cdot \rho R^3}{4R^2} = \frac{16}{3} \cdot \frac{G \cdot \pi \cdot \rho R^3}{R^2} = \frac{16}{3} g

We find that the new gravitational field strength is:
g′=4gg' = 4g
Therefore, the correct answer is B) 4g4g.

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