Define gravitational potential at a point. 0 3 . 2 Figure 2 shows the positions of equipotential surfaces at different distances from the centre of the Moon. Explain how the equipotential surfaces in Figure 2 show that the gravitational field is not uniform. 0 3 . 3 Calculate, using Figure 2, the escape velocity at the surface of the Moon. radius of Moon = 1.74 × 10^6 m

A-Level AQA Physics Gravitational Potential: Define gravitational potential a

Define gravitational potential at a point - AQA - A-Level Physics - Question 3 - 2019 - Paper 2

Question 3

Define gravitational potential at a point. 0 3 . 2 Figure 2 shows the positions of equipotential surfaces at different distances from the centre of the Moon. Exp... show full transcript

Worked Solution & Example Answer:Define gravitational potential at a point - AQA - A-Level Physics - Question 3 - 2019 - Paper 2

Step 1

Define gravitational potential at a point.

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Answer

Gravitational potential at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point against the gravitational field.

Step 2

Explain how the equipotential surfaces in Figure 2 show that the gravitational field is not uniform.

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Answer

The equipotential surfaces in Figure 2 show variations in gravitational potential at different distances from the Moon's center. As the distance from the center of the Moon increases, the spacing between the equipotential surfaces decreases, indicating that the gravitational field strength changes. The non-parallel lines signify that the gravitational force is stronger closer to the Moon and weaker further away.

Step 3

Calculate, using Figure 2, the escape velocity at the surface of the Moon.

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Answer

To calculate the escape velocity ( $v_e$ ) at the surface of the Moon, we use the formula: $v_e = \sqrt{2gh}$ where:

$g$ is the gravitational acceleration at the surface, and
$h$ is the height, which is equivalent to the radius of the Moon.

From Figure 2, we can infer $g$ using the potential difference between equipotential surfaces: $g = -\frac{ΔV}{Δr}$ Using the values: $r = 1.74 × 10^6 ext{ m}$ we find: $v_e = \sqrt{2 imes 2.4 × 10^2 ext{ m/s}^{2} \times 1.74 × 10^6 ext{ m}}$ After calculating, $v_e ≈ 2.3 \times 10^{3} ext{ m/s}$ .

Define gravitational potential at a point - AQA - A-Level Physics - Question 3 - 2019 - Paper 2

Worked Solution & Example Answer:Define gravitational potential at a point - AQA - A-Level Physics - Question 3 - 2019 - Paper 2

Define gravitational potential at a point.

Explain how the equipotential surfaces in Figure 2 show that the gravitational field is not uniform.

Calculate, using Figure 2, the escape velocity at the surface of the Moon.

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