Phase 3 Consolidation

MC1. B — $g' = g(R/(R+h))^2 = g(R/3R)^2 = g/9$. Error A: uses $g = 9.8$ constant. Error C: forgets $r = R + h$, uses $h = 2R$ so $r = 2R$ giving $g/4$.

MC2. C — $U = -GMm/r$ with the mandatory negative sign. Error A: forgets negative. Error B: uses $r^2$ instead of $r$.

MC3. A — $v_e' = \sqrt{2G(2M)/(2R)} = \sqrt{2GM/R} = v_e$. The changes cancel. Error B: only considers mass doubling.

MC4. D — $E = K + U = \frac{1}{2}mv^2 - GMm/r = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$. Negative and half the magnitude of $U$.

MC5. B — Newton showed this is valid for all closed orbits using semi-major axis $a$. Error A confuses it with circular-only approximations.

SA1. (a) Equating $mv^2/r = GMm/r^2$: $v = \sqrt{GM_E/r} = \sqrt{(6.67 \times 10^{-11})(5.97 \times 10^{24})/(1.0 \times 10^7)} = \sqrt{3.98 \times 10^7} = \boxed{6.31 \times 10^3 \text{ m/s}}$

(b) $K = \frac{1}{2}mv^2 = 0.5 \times 1500 \times (6.31 \times 10^3)^2 = \boxed{2.99 \times 10^{10} \text{ J}}$ (or $3.0 \times 10^{10}$ J)

(c) $E_{\text{total}} = -K = \boxed{-2.99 \times 10^{10} \text{ J}}$ (or use $E = -GMm/2r$ directly)

SA2. $V = -GM/r$ is defined as the work done per unit mass to bring a test mass from infinity to distance $r$. Since gravity is attractive, this work is done by the field, not against it, so the potential is negative. At $r = \infty$, no work is done by or against the field, so $V = 0$. The zero is chosen at infinity because it is the only natural reference point where the gravitational influence of the mass $M$ becomes zero. The negative sign reflects that any finite distance requires negative work to remove the mass to infinity (i.e., the mass is bound).

SA3. The statement is partially correct but misleading.

Both $g = GM/r^2$ and $V = -GM/r$ do decrease in magnitude as $r$ increases. A more negative $V$ does correspond to a smaller $r$, and at smaller $r$, $g$ is indeed larger.

However, the relationship is not direct: $g \propto 1/r^2$ while $V \propto 1/r$. For example:

At $r = R_E = 6.37 \times 10^6$ m: $V = -6.25 \times 10^7$ J/kg and $g = 9.81$ m/s$^2$.

At $r = 2R_E = 1.274 \times 10^7$ m: $V = -3.13 \times 10^7$ J/kg and $g = 2.45$ m/s$^2$.

When $r$ doubles, $V$ halves (becomes half as negative) but $g$ quarters. So while the trend is the same, the quantitative relationship differs. The statement ignores the different power laws ($1/r$ vs $1/r^2$) and the fact that $g = |dV/dr|$, not $g \propto V$.

MQ1. $F = \dfrac{Gm_1m_2}{r^2} = \dfrac{(6.67 \times 10^{-11})(50)(50)}{(0.50)^2} = \dfrac{1.67 \times 10^{-7}}{0.25} = \boxed{6.7 \times 10^{-7} \text{ N}}$

MQ2. $r = R_E + h = 6.37 \times 10^6 + 2.0 \times 10^6 = 8.37 \times 10^6$ m

$g = \dfrac{GM_E}{r^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(8.37 \times 10^6)^2} = \dfrac{3.98 \times 10^{14}}{7.01 \times 10^{13}} = \boxed{5.68 \text{ m/s}^2}$

(Or using ratio: $g' = 9.81 \times (6.37/8.37)^2 = 9.81 \times 0.579 = 5.68$ m/s$^2$)

MQ3. The zero-reference point for gravitational potential energy is at $r = \infty$ (infinite separation).

This point is chosen because: (1) it is the only natural, unambiguous reference where the gravitational force between two masses becomes exactly zero; (2) it ensures that all bound systems have negative total energy, making it easy to identify bound vs. unbound orbits; (3) the expression $U = -GMm/r$ then correctly represents the work needed to separate the masses to infinity.

MQ4. $K = \dfrac{GM_Em}{2r} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(2000)}{2 \times 8.5 \times 10^6} = \dfrac{7.96 \times 10^{17}}{1.70 \times 10^7} = \boxed{4.68 \times 10^{10} \text{ J}}$

$U = -2K = \boxed{-9.36 \times 10^{10} \text{ J}}$

$E_{\text{total}} = K + U = \boxed{-4.68 \times 10^{10} \text{ J}}$

MQ5. Moon: $v_e = \sqrt{\dfrac{2GM_M}{R_M}} = \sqrt{\dfrac{2(6.67 \times 10^{-11})(7.35 \times 10^{22})}{1.74 \times 10^6}} = \sqrt{5.63 \times 10^6} = \boxed{2.37 \times 10^3 \text{ m/s} = 2.37 \text{ km/s}}$

Earth: $v_e = \sqrt{\dfrac{2(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.37 \times 10^6}} = \boxed{1.12 \times 10^4 \text{ m/s} = 11.2 \text{ km/s}}$

Ratio: $v_{e,\text{Moon}} / v_{e,\text{Earth}} = 2.37/11.2 = 0.21$, or about one-fifth of Earth's escape velocity.

MQ6. $V = -\dfrac{GM_E}{r} = -\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{2.0 \times 10^7} = \boxed{-1.99 \times 10^7 \text{ J/kg}}$ (or $-2.0 \times 10^7$ J/kg)

$g = \dfrac{GM_E}{r^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(2.0 \times 10^7)^2} = \dfrac{3.98 \times 10^{14}}{4.0 \times 10^{14}} = \boxed{0.995 \text{ m/s}^2 \approx 1.0 \text{ m/s}^2}$

MQ7. $W = \Delta U = GM_Em\left(\dfrac{1}{R_E} - \dfrac{1}{r_2}\right)$

$W = (6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)\left(\dfrac{1}{6.37 \times 10^6} - \dfrac{1}{1.5 \times 10^7}\right)$

$W = 3.98 \times 10^{17} \times (1.570 \times 10^{-7} - 6.667 \times 10^{-8})$

$W = 3.98 \times 10^{17} \times 9.03 \times 10^{-8} = \boxed{3.59 \times 10^{10} \text{ J}}$ (accept $3.5$-$3.7 \times 10^{10}$ J)

MQ8. Starting from $V = -\dfrac{GM}{r} = -GM \cdot r^{-1}$:

$\dfrac{dV}{dr} = -GM \cdot (-1) \cdot r^{-2} = \dfrac{GM}{r^2}$

Therefore $g = -\dfrac{dV}{dr} = -\dfrac{GM}{r^2}$

But wait: $g$ is the magnitude of the field strength, and the field points inward (toward decreasing $r$). The gravitational field vector is $\vec{g} = -\dfrac{GM}{r^2}\hat{r}$, so $g = -\dfrac{dV}{dr}$ correctly gives the radial component with its sign. The negative sign indicates that $V$ increases (becomes less negative) as $r$ increases, so the field points in the direction of decreasing $r$.

MQ9. $M_P = 4M_E$, $R_P = 2R_E$

$g_P = \dfrac{GM_P}{R_P^2} = \dfrac{G(4M_E)}{(2R_E)^2} = \dfrac{4GM_E}{4R_E^2} = \dfrac{GM_E}{R_E^2} = g_E$

So $g_P = \boxed{1 \times g_E}$ (the same as Earth's surface gravity)

$v_{eP} = \sqrt{\dfrac{2GM_P}{R_P}} = \sqrt{\dfrac{2G(4M_E)}{2R_E}} = \sqrt{2} \times \sqrt{\dfrac{2GM_E}{R_E}} = \sqrt{2} \times v_{eE}$

So $v_{eP} = \boxed{\sqrt{2} \times v_{eE} \approx 1.41 \times v_{eE}}$

MQ10. $T = 1.77$ days $= 1.77 \times 24 \times 3600 = 1.529 \times 10^5$ s

Using $T^2 = \dfrac{4\pi^2}{GM_J}\,r^3$:

$r^3 = \dfrac{GM_JT^2}{4\pi^2} = \dfrac{(6.67 \times 10^{-11})(1.90 \times 10^{27})(1.529 \times 10^5)^2}{4\pi^2}$

$r^3 = \dfrac{(6.67 \times 10^{-11})(1.90 \times 10^{27})(2.338 \times 10^{10})}{39.478} = \dfrac{2.96 \times 10^{27}}{39.478} = 7.50 \times 10^{25}$

$r = \sqrt[3]{7.50 \times 10^{25}} = \boxed{4.22 \times 10^8 \text{ m}}$ (accept $4.1$-$4.3 \times 10^8$ m)

$v = \dfrac{2\pi r}{T} = \dfrac{2\pi(4.22 \times 10^8)}{1.529 \times 10^5} = \boxed{1.73 \times 10^4 \text{ m/s} = 17.3 \text{ km/s}}$

TQ11. (a) For circular orbit: $E_{\text{total}} = -\dfrac{GM_Em}{2r}$

$E = -\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(500)}{2(9.0 \times 10^6)} = -\dfrac{1.99 \times 10^{17}}{1.8 \times 10^7} = \boxed{-1.11 \times 10^{10} \text{ J}}$ [2 marks: formula + substitution 1 mark, answer 1 mark]

(b) To escape, total energy must be $\geq 0$. Minimum additional energy $= |E_{\text{total}}| = \boxed{1.11 \times 10^{10} \text{ J}}$ [2 marks: reasoning 1 mark, answer 1 mark]

TQ12. (a) $V = -\dfrac{GM_E}{r} = -\dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{1.2 \times 10^7} = \boxed{-3.32 \times 10^7 \text{ J/kg}}$

$g = \dfrac{GM_E}{r^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(1.2 \times 10^7)^2} = \boxed{2.77 \text{ m/s}^2}$ [2 marks]

(b) Equipotential surfaces are spherical because $V = -GM/r$ depends only on $r$, not direction. Any point at the same distance $r$ from the centre has the same potential. Field lines are radial and perpendicular to equipotentials, pointing inward. [concentric circles for equipotentials, radial inward arrows for field lines, labelled $V_1 > V_2 > V_3$ (less negative further out).] [2 marks]

TQ13. (a) $T = 45$ days $= 45 \times 24 \times 3600 = 3.888 \times 10^6$ s

$r^3 = \dfrac{GM_{\star}T^2}{4\pi^2} = \dfrac{(6.67 \times 10^{-11})(2.0 \times 10^{30})(3.888 \times 10^6)^2}{4\pi^2}$

$r^3 = \dfrac{(6.67 \times 10^{-11})(2.0 \times 10^{30})(1.512 \times 10^{13})}{39.478} = \dfrac{2.017 \times 10^{33}}{39.478} = 5.11 \times 10^{31}$

$r = \sqrt[3]{5.11 \times 10^{31}} = \boxed{3.71 \times 10^{10} \text{ m}}$ [2 marks]

(b) The claim is incorrect. If $M_{\star}$ doubles, Kepler's Third Law tells us $r$ must change for the same period. More directly, $v = \sqrt{GM/r}$: if $M$ doubles while $r$ stays constant, $v$ would increase by $\sqrt{2}$. However, in reality, the orbital parameters would adjust. The student confuses maintaining the same orbit with maintaining the same conditions. If the planet stayed at the same $r$ with a doubled central mass, $v$ would indeed change, and the orbit would no longer be stable at the original radius. [2 marks]