A-Level Physics May/June 2025 Q1(b)(i): Mars is a planet that may be considered to be an isolated uniform sphere of radius 3.4…

1 answer

• ▲0
  accepted ✓

  The change in gravitational potential energy ($ΔE_p$) is given by the difference between the final and initial potential energies.

  $E_p = -\frac{GMm}{r}$

  $\Delta E_p = E_{p,final} - E_{p,initial} = \left(-\frac{GMm}{r_2}\right) - \left(-\frac{GMm}{r_1}\right)$

  $\Delta E_p = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$

  First, calculate the initial and final orbital radii ($r_1$ and $r_2$) from the center of Mars.

  Initial radius $r_1 = \text{radius of Mars} + \text{initial height}$ $r_1 = (3.4 \times 10^6) + (1.7 \times 10^6) = 5.1 \times 10^6$ m

  Final radius $r_2 = \text{radius of Mars} + \text{final height}$ $r_2 = (3.4 \times 10^6) + (6.8 \times 10^6) = 10.2 \times 10^6$ m

  Now substitute the known values into the equation for $\Delta E_p$:

  $5.1 \times 10^8 = (6.67 \times 10^{-11}) \times M \times 122 \times \left(\frac{1}{5.1 \times 10^6} - \frac{1}{10.2 \times 10^6}\right)$

  $5.1 \times 10^8 = (8.1374 \times 10^{-9}) \times M \times (1.9608 \times 10^{-7} - 9.8039 \times 10^{-8})$

  $5.1 \times 10^8 = (8.1374 \times 10^{-9}) \times M \times (9.804 \times 10^{-8})$

  Rearrange to solve for M, the mass of Mars:

  $M = \frac{5.1 \times 10^8}{(8.1374 \times 10^{-9}) \times (9.804 \times 10^{-8})}$

  $M = \frac{5.1 \times 10^8}{7.978 \times 10^{-16}}$

  $M = 6.392 \times 10^{23}$ kg

  To two significant figures, the mass of Mars is $6.4 \times 10^{23}$ kg.

  ---

  How the marks are awarded

  • C1 — The first mark is awarded for correctly calculating the orbital radii by adding the radius of Mars to the height of the satellite above the surface. In the model answer, this is shown by the calculation of r1 = 5.1 x 10^6 m and r2 = 10.2 x 10^6 m.
  • C1 — The second mark is for correctly setting up the expression for the change in gravitational potential energy. This involves substituting the satellite mass (122 kg) and at least one of the correct radii into the formula, as shown by the line: $5.1 \times 10^8 = G \times M \times 122 \times (\frac{1}{r_1} - \frac{1}{r_2})$.
  • A1 — The final mark is awarded for the correct substitution of all values into the full equation, followed by a correct calculation that leads to the value of M = 6.4 x 10^23 kg (when rounded to 2 significant figures).

  Common mistakes

  • Using the height above the surface (e.g., 1.7 x 10^6 m) in the GPE formula instead of the orbital radius from the planet's center (5.1 x 10^6 m).
  • Making a sign error when calculating the change in potential energy, for example by calculating $E_{p,initial} - E_{p,final}$ or forgetting that GPE is a negative quantity.
  • Incorrectly applying the formula for potential energy change near a surface, $\Delta E_p = mgh$, which is only valid in a uniform gravitational field.
  • Errors in rearranging the equation to solve for M, or calculator input errors, especially with the inverse terms and powers of ten.

  Examiner tip: Always remember that for orbital mechanics and gravitational fields, distances (radii) are measured from the center of the central mass, not from its surface.

  ---

  AI-generated model answer, grounded in the official Cambridge mark scheme and reviewed by the MarkScheme team. Mark your own answer to this question →

  @markscheme_answers

Your answer

Sign in to answer this question.