Checkpoint 3: Gravitational Fields

Multiple Choice Answers

1. C — $F = \dfrac{(6.67 \times 10^{-11}) \times 5 \times 5}{(0.5)^2} = \dfrac{1.6675 \times 10^{-9}}{0.25} = 6.67 \times 10^{-9}$ N. A uses $r = 1$ m, C is $1.33 \times 10^{-9}$ (forgot to square $r$), D uses $r = 0.25$ m.

2. C — From $F \propto 1/r^2$, halving $r$ gives $F' = F/(1/2)^2 = 4F$. A mistakes inverse square for inverse proportionality, B ignores distance, D has the change backwards.

3. C — $g = g_0 \times (R/r)^2 = 9.8 \times (1/3)^2 = 9.8/9 = 1.09$ m/s$^2$. A is surface value, B divides by 3 instead of 9, D divides by 27.

4. A — Gravitational force is always attractive, acting along the line joining centres. B and C describe repulsion which never occurs, D describes a perpendicular force.

5. B — $U = -GMm/r$ with negative sign mandatory; $U = 0$ at $r = \infty$. A has wrong sign, C has wrong dependence, D uses $1/r^2$ like force.

6. A — Gravity points inward; displacement is outward, so $W = \vec{F} \cdot \vec{d} < 0$. Gravity does negative work as the satellite moves outward. B confuses altitude gain with work sign, C confuses conservative with zero work.

7. B — Gravitational potential is GPE per unit mass: $V = U/m = -GM/r$ (J/kg). A confuses potential with field strength, C and D mix in kinetic energy.

8. B — The negative sign means $g$ points toward decreasing $V$ (toward the mass). $|g| = |dV/dr|$ gives the magnitude. A reverses the relationship, C claims direct proportionality which is false, D ignores the vector nature.

9. B — Equipotential surfaces have constant $V$; since $\Delta U = q\Delta V = 0$ along them, no work is done. A confuses equipotential with equal field strength, C is only true for uniform fields, D confuses potential with kinetic energy.

10. C — $v_e = \sqrt{2GM/r}$: depends only on the planet's $M$ and $r$. The escaping object's mass cancels out. A and B incorrectly include the object's mass, D confuses escape velocity with orbital launch.

11. A — $R_s = 2GM/c^2$ is the event horizon radius. B confuses it with the singularity (a point, $r = 0$), C suggests infinite force at $R_s$ which is wrong, D describes the photon sphere.

12. C — Kepler's Third Law: $T^2 \propto r^3$. A and B have wrong exponents, D inverts the relationship.

13. B — For circular orbits: $K = +GMm/2r$, $U = -GMm/r$, so $K = -U/2$ and $E = K + U = -GMm/2r = -K = U/2$. A, C, D have incorrect energy relationships.

14. B — Moving to higher orbit: $U$ increases (becomes less negative), so work done against gravity is positive. A has the sign wrong, C confuses orbital energy conservation with work done, D ignores potential energy change.

15. B — $\dfrac{T_X^2}{T_Y^2} = \dfrac{(4r)^3}{(9r)^3} = \dfrac{64}{729}$, so $\dfrac{T_X}{T_Y} = \sqrt{\dfrac{64}{729}} = \dfrac{8}{27}$. A uses $T \propto r$, C uses $T \propto r^2$, D takes $\sqrt{2/3}$.

Short Answer 1 (3 marks) — Gravitational Force and Field Strength at Altitude

Step 1: Find the centre-to-centre distance: $r = R_E + h = 6.37 \times 10^6 + 6.00 \times 10^5 = 6.97 \times 10^6$ m (1 mark for correct $r$)

Step 2: Calculate the gravitational force:

$F = \dfrac{GM_E m}{r^2} = \dfrac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 1000}{(6.97 \times 10^6)^2}$

$F = \dfrac{3.98 \times 10^{17}}{4.86 \times 10^{13}} = 8.19 \times 10^3$ N (1 mark)

Step 3: Calculate gravitational field strength:

$g = \dfrac{GM_E}{r^2} = \dfrac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24})}{(6.97 \times 10^6)^2} = \dfrac{3.98 \times 10^{14}}{4.86 \times 10^{13}} = 8.19$ N/kg (or m/s$^2$) (1 mark)

Alternatively: $g = F/m = 8.19 \times 10^3/1000 = 8.19$ N/kg. This is about 84% of surface gravity — astronauts at this altitude still experience strong gravity.

Short Answer 2 (3 marks) — Change in GPE and Work Done

Step 1: Calculate GPE at each position:

$U_1 = -\dfrac{GM_E m}{r_1} = -\dfrac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 500}{7.0 \times 10^6} = -\dfrac{1.99 \times 10^{14}}{7.0 \times 10^6} = -2.84 \times 10^7$ J

$U_2 = -\dfrac{GM_E m}{r_2} = -\dfrac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 500}{1.2 \times 10^7} = -\dfrac{1.99 \times 10^{14}}{1.2 \times 10^7} = -1.66 \times 10^7$ J (1 mark for both GPE values)

Step 2: Calculate the change in GPE:

$\Delta U = U_2 - U_1 = (-1.66 \times 10^7) - (-2.84 \times 10^7) = +1.18 \times 10^7$ J (1 mark)

Step 3: The work done by the gravitational field is:

$W_{\text{field}} = -\Delta U = -1.18 \times 10^7$ J (1 mark)

Alternatively using $W = GMm(1/r_1 - 1/r_2)$:

$W = (6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 500 \times \left(\dfrac{1}{7.0 \times 10^6} - \dfrac{1}{1.2 \times 10^7}\right)$

$W = 1.99 \times 10^{14} \times (1.429 \times 10^{-7} - 8.333 \times 10^{-8}) = 1.99 \times 10^{14} \times 5.95 \times 10^{-8} = 1.18 \times 10^7$ J (work done against gravity)

The gravitational field does negative work as the probe moves outward (force and displacement are in opposite directions).

Short Answer 3 (4 marks) — Potential Gradient and Field Strength

Step 1: Convert values and find the potential difference:

$V_1 = -45$ MJ/kg $= -45 \times 10^6$ J/kg, $V_2 = -36$ MJ/kg $= -36 \times 10^6$ J/kg

$\Delta V = V_2 - V_1 = (-36 \times 10^6) - (-45 \times 10^6) = +9 \times 10^6$ J/kg

$\Delta r = 3000$ km $= 3.0 \times 10^6$ m (1 mark for correct conversions)

Step 2: Estimate average field strength:

$g_{\text{avg}} = -\dfrac{\Delta V}{\Delta r} = -\dfrac{9 \times 10^6}{3.0 \times 10^6} = -3.0$ N/kg (1 mark)

The magnitude is $|g| = 3.0$ N/kg (or 3.0 m/s$^2$). (1 mark)

Step 3: The negative sign in $g = -dV/dr$ means:

Gravitational potential $V$ increases (becomes less negative) as $r$ increases. Since $V$ increases with $r$, the gradient $dV/dr$ is positive. The negative sign ensures that $g$ points in the direction of decreasing $r$ — that is, toward the centre of the Earth. The field strength vector points inward, toward the mass that creates the field. This convention ensures that $g$ correctly represents the direction of gravitational force on a test mass. (1 mark for explanation)

Short Answer 4 (4 marks) — Escape Velocity and Schwarzschild Radius for Mars

Step 1 — Escape velocity:

$v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{2 \times (6.67 \times 10^{-11}) \times (6.42 \times 10^{23})}{3.40 \times 10^6}}$ (1 mark for formula and substitution)

$v_e = \sqrt{\dfrac{8.56 \times 10^{13}}{3.40 \times 10^6}} = \sqrt{2.52 \times 10^7} = 5.02 \times 10^3$ m/s $= 5.02$ km/s (1 mark)

Step 2 — Schwarzschild radius:

$R_s = \dfrac{2GM}{c^2} = \dfrac{2 \times (6.67 \times 10^{-11}) \times (6.42 \times 10^{23})}{(3.0 \times 10^8)^2}$ (1 mark)

$R_s = \dfrac{8.56 \times 10^{13}}{9.0 \times 10^{16}} = 9.51 \times 10^{-4}$ m $\approx 0.95$ mm

Step 3 — Discussion:

Mars cannot become a black hole under any known physical process. The Schwarzschild radius (~0.95 mm) is many orders of magnitude smaller than Mars's actual radius (3.4 million metres). For Mars to become a black hole, all its mass would need to be compressed into a sphere less than 1 mm across — an impossible density ($\rho \approx 10^{25}$ kg/m$^3$) far exceeding anything achievable by gravitational collapse. Stellar-mass black holes form from stars with core masses exceeding the Chandrasekhar limit (~1.4 solar masses) that undergo supernova collapse. Mars's mass ($\sim 0.1\%$ of solar mass) is far below any threshold for gravitational collapse to a black hole. (1 mark for discussion)

Short Answer 5 (4 marks) — Kepler's Third Law for Europa

Step 1: Convert the orbital period to seconds:

$T = 3.55 \times 24 \times 3600 = 3.067 \times 10^5$ s (1 mark for conversion)

Step 2: Rearrange Kepler's Third Law:

$T^2 = \dfrac{4\pi^2}{GM_J} r^3 \implies r^3 = \dfrac{GM_J T^2}{4\pi^2}$

$r^3 = \dfrac{(6.67 \times 10^{-11}) \times (1.90 \times 10^{27}) \times (3.067 \times 10^5)^2}{4\pi^2}$ (1 mark for correct substitution)

$r^3 = \dfrac{(6.67 \times 10^{-11}) \times (1.90 \times 10^{27}) \times 9.41 \times 10^{10}}{39.48}$

$r^3 = \dfrac{1.19 \times 10^{28}}{39.48} = 3.02 \times 10^{26}$ m$^3$

Step 3: Calculate $r$:

$r = \sqrt[3]{3.02 \times 10^{26}} = 6.71 \times 10^8$ m (1 mark)

$r = 6.71 \times 10^8$ m $= 6.71 \times 10^5$ km $= 671,000$ km

Step 4 — Verification:

The calculated orbital radius of 671,000 km is in excellent agreement with the known value of 671,000 km (to 3 significant figures). This confirms that Kepler's Third Law, when combined with Newton's law of gravitation, accurately predicts the orbital parameters of moons around planets. The small difference ($< 0.1\%$) arises from rounding in the given values. (1 mark for verification)