Gravitational potential and comparison with electric fieldsEdexcel A-Level Physics: Subtopic test

1. 1
   An engineer is planning to launch a probe of mass 450 kg from the surface of the Earth. The Earth may be treated as a uniform sphere of mass 5.97×10²⁴ kg and radius 6.37×10⁶ m, and the gravitational constant is G = 6.67×10⁻¹¹ N m² kg⁻².

   (a)

   The gravitational potential at the surface of the Earth is negative. Which statement gives the reason?

   [1 mark]

   • AIt is defined as zero at infinity, and work must be done on a mass to move it from the surface to infinity because gravity is attractive
   • BGravity is a repulsive force at the surface of the Earth
   • CThe mass of the Earth is a negative quantity in the equation V = −GM/r
   • DThe potential has the same direction as the force, which is downwards

   (b)

   What is the gravitational potential at the surface of the Earth?

   [1 mark]

   • A+6.25×10⁷ J kg⁻¹
   • B−9.81 J kg⁻¹
   • C−6.25×10⁷ J kg⁻¹
   • D−3.98×10¹⁴ J kg⁻¹

   (c)

   The probe is raised from the surface of the Earth to a height equal to the radius of the Earth. Calculate the increase in the gravitational potential energy of the probe.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   In a simple model of a hydrogen atom, an electron of mass 9.11×10⁻³¹ kg and charge −1.60×10⁻¹⁹ C is a distance of 5.3×10⁻¹¹ m from a proton of mass 1.67×10⁻²⁷ kg and charge +1.60×10⁻¹⁹ C. Both particles are treated as point masses and point charges. Take G = 6.67×10⁻¹¹ N m² kg⁻² and ε₀ = 8.85×10⁻¹² F m⁻¹.

   (a)

   Which feature is shared by Newton's law of gravitation and Coulomb's law?

   [1 mark]

   • AThe force depends on the product of two masses
   • BThe force is always attractive
   • CBoth forces can be repulsive or attractive
   • DThe force is inversely proportional to the square of the separation of the particles

   (b)

   The separation of the proton and the electron is doubled. What happens to the ratio (electric force) / (gravitational force) between them?

   [1 mark]

   • AIt halves
   • BIt stays the same
   • CIt doubles
   • DIt becomes four times larger

   (c)

   Calculate the gravitational force between the proton and the electron.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   A lunar lander of mass 1200 kg is on the surface of the Moon. The Moon may be treated as a uniform sphere of mass 7.35×10²² kg and radius 1.74×10⁶ m. Take G = 6.67×10⁻¹¹ N m² kg⁻². The surface field strength of the Moon is 1.62 N kg⁻¹.

   (a)

   Calculate the gravitational potential at the surface of the Moon and explain why the value is negative.

   [3 marks]

   (b)

   The lander is raised to a height of 1.74×10⁶ m above the surface of the Moon. Calculate the gain in gravitational potential energy, and explain why using ΔE = mgh with g = 1.62 N kg⁻¹ would give a different answer.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   Two identical small spheres, each of mass 0.50 kg and carrying a charge of +1.0 µC, have their centres 0.20 m apart in a vacuum. A student is writing a summary that compares the gravitational field around a planet with the electric field around a charged object. Take G = 6.67×10⁻¹¹ N m² kg⁻² and ε₀ = 8.85×10⁻¹² F m⁻¹.

   (a)

   Compare the gravitational field around an isolated point mass with the electric field around an isolated point charge, referring to the force laws, the nature of the force and the potentials.

   [6 marks]

   (b)

   Calculate the gravitational force and the electric force between the two spheres, and use your results to evaluate why gravity is usually ignored when charged objects interact but is dominant for planets and stars.

   [6 marks]

   Total for question 4: 12 marks

End of questions