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Cambridge International AS & A Level · 13 · structured

A spherical planet: field, potential and energy

A spherical planet: field, potential and energy · Original GioPhysics question with a detailed, mark-by-mark answer guide.

Demand
demanding
Marks
11
Topics
6
Answer
Complete
structured11 marks

A uniform spherical planet has radius R=8.0×10⁶ m and GM=4.0×10¹⁴ m³ s⁻². A small object of mass 2.0×10³ kg can be moved slowly outside it. Neglect other gravitating bodies.

  1. (a)

    Explain why the exterior field can be calculated using a point mass and state where that point mass is located. Derive g=GM/r² from gravitational force and the definition of g.

    3 marks
  2. (b)

    Explain why g is approximately constant over a 100 m change of height near the surface without claiming that it is exactly constant.

    2 marks
  3. (c)

    Define gravitational potential with the infinity reference and calculate the surface value.

    3 marks
  4. (d)

    Calculate the object's gravitational potential energy at R and at 2R, and the external work needed to move it slowly between those positions. Explain the sign.

    3 marks
Ready to self-mark?Reveal the detailed answer guide
Answer overview(a) For r≥R the uniform sphere acts as mass at its centre; g=F/m=GM/r². (b) 100 m is tiny compared with R; (R/(R+100))² is nearly 1. (c) External work per unit small test mass brought slowly from infinity; φ(R)=−5.0×10⁷ J kg⁻¹. (d) E_p(R)=−1.0×10¹¹ J; E_p(2R)=−5.0×10¹⁰ J; W_ext=+5.0×10¹⁰ J.
01

(a)

3 marks

Explain why the exterior field can be calculated using a point mass and state where that point mass is located. Derive g=GM/r² from gravitational force and the definition of g.

How to approach it

Identify what the question is testing, organise the response into distinct mark-earning points, and make every conclusion traceable to a physical principle or to the evidence supplied.

  1. 1

    Use the spherical symmetry result only outside the sphere: replace its mass by a point at its centre.

  2. 2

    Newton's force magnitude is F=GMm/r².

  3. 3

    Divide by test mass: g=F/m=GM/r².

02

(b)

2 marks

Explain why g is approximately constant over a 100 m change of height near the surface without claiming that it is exactly constant.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    The relevant distance changes from R to R+100 m, a fractional change 100/(8.0×10⁶)=1.25×10⁻⁵.

  2. 2

    By inverse-square dependence the fractional field change is only about 2.5×10⁻⁵, so local constancy is a good approximation.

03

(c)

3 marks

Define gravitational potential with the infinity reference and calculate the surface value.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    Define the signed external work per unit small test mass bringing it from infinity with no kinetic-energy change.

  2. 2

    Choose φ=0 at infinity and use φ=−GM/r.

  3. 3

    φ(R)=−4.0×10¹⁴/(8.0×10⁶)=−5.0×10⁷ J kg⁻¹.

04

(d)

3 marks

Calculate the object's gravitational potential energy at R and at 2R, and the external work needed to move it slowly between those positions. Explain the sign.

How to approach it

List the given quantities with units, identify the required quantity, write the governing relationship before substituting, and keep extra digits until the final line so rounding does not distort the result.

  1. 1

    Pair potential energy is E_p=mφ=−GMm/r.

  2. 2

    At R this is −1.0×10¹¹ J; at 2R it is −5.0×10¹⁰ J.

  3. 3

    The increase is +5.0×10¹⁰ J. The external agent works against attraction; positive energy is required to separate the masses.

Complete marking guidance

Use these worked steps alongside the part-by-part answers. They explain the method and mark allocation; they do not add extra marks.

  1. (a) Use the spherical symmetry result only outside the sphere: replace its mass by a point at its centre.
  2. (a) Newton's force magnitude is F=GMm/r².
  3. (a) Divide by test mass: g=F/m=GM/r².
  4. (b) The relevant distance changes from R to R+100 m, a fractional change 100/(8.0×10⁶)=1.25×10⁻⁵.
  5. (b) By inverse-square dependence the fractional field change is only about 2.5×10⁻⁵, so local constancy is a good approximation.
  6. (c) Define the signed external work per unit small test mass bringing it from infinity with no kinetic-energy change.
  7. (c) Choose φ=0 at infinity and use φ=−GM/r.
  8. (c) φ(R)=−4.0×10¹⁴/(8.0×10⁶)=−5.0×10⁷ J kg⁻¹.
  9. (d) Pair potential energy is E_p=mφ=−GMm/r.
  10. (d) At R this is −1.0×10¹¹ J; at 2R it is −5.0×10¹⁰ J.
  11. (d) The increase is +5.0×10¹⁰ J. The external agent works against attraction; positive energy is required to separate the masses.

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