A2 Gravitational FieldsNuclear Physics 5.4.1(b)5.4.2(a)5.4.2(b)5.4.3(b)5.4.3(a)

~25 min Difficulty: 7/10

Prior knowledge Volume of a spherebasic circular motionNewton's Law of Gravitation.

Problem structure

(a) 2, numerical, density/volume/mass relationship. (b) 2, numerical, gravitational field strength calculation. (c)(i) 4, numerical, Kepler's 3rd Law derivation/application. (c)(ii) 2, numerical, orbital velocity calculation. (d) 4, numerical, ratio reasoning and orbital geometry.

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A hypothetical "Nuclear Star" is a dense, spherical object composed entirely of neutrons. It has a radius $R$ of $12.0\text{ km}$ and a uniform density equal to that of an atomic nucleus, $\rho = 2.30 \times 10^{17}\text{ kg m}^{-3}$. (a) Show that the mass $M$ of the Nuclear Star is approximately $1.7 \times 10^{30}\text{ kg}$. [2] (b) Calculate the gravitational field strength $g$ at the surface of the star. [2] (c) A small satellite is in a stable circular orbit at a height of $4.0\text{ km}$ above the surface of the star. (i) Calculate the orbital period $T$ of the satellite. [4] (ii) Calculate the orbital speed $v$ of the satellite. [2] (d) A second satellite is placed in a higher circular orbit such that its orbital period is exactly double that of the first satellite. Determine the height $h_2$ of this second satellite above the surface of the star. [4] Data: $G = 6.67 \times 10^{-11}\text{ N m}^2\text{ kg}^{-2}$

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(a) The star is a sphere. The volume $V$ is given by: $V = \frac{4}{3}\pi R^3$ $V = \frac{4}{3}\pi (12.0 \times 10^3)^3 = 7.238 \times 10^{12}\text{ m}^3$ Mass $M = \rho V = (2.30 \times 10^{17})(7.238 \times 10^{12}) = 1.6648 \times 10^{30}\text{ kg}$ $M \approx 1.7 \times 10^{30}\text{ kg}$ (as required). (b) Using the formula for gravitational field strength at the surface: $g = \frac{GM}{R^2}$ $g = \frac{(6.67 \times 10^{-11})(1.6648 \times 10^{30})}{(12.0 \times 10^3)^2}$ $g = \frac{1.110 \times 10^{20}}{1.44 \times 10^8} = 7.71 \times 10^{11}\text{ N kg}^{-1}$ (or $\text{m s}^{-2}$) (c) (i) For a circular orbit, the gravitational force provides the centripetal force: $\frac{GMm}{r^2} = m\omega^2 r$ where $r = R + h = 12.0 + 4.0 = 16.0\text{ km} = 1.60 \times 10^4\text{ m}$. $\omega^2 = \frac{GM}{r^3}$ Since $\omega = \frac{2\pi}{T}$, then $T^2 = \frac{4\pi^2 r^3}{GM}$ (Kepler's 3rd Law). $T = \sqrt{\frac{4\pi^2 (1.60 \times 10^4)^3}{(6.67 \times 10^{-11})(1.6648 \times 10^{30})}}$ $T = \sqrt{\frac{1.617 \times 10^{14}}{1.110 \times 10^{20}}} = \sqrt{1.456 \times 10^{-6}}$ $T = 1.207 \times 10^{-3}\text{ s} = 1.21\text{ ms}$ (to 3 SF). (ii) Orbital speed $v = \frac{2\pi r}{T}$ or $v = \sqrt{\frac{GM}{r}}$ $v = \sqrt{\frac{(6.67 \times 10^{-11})(1.6648 \times 10^{30})}{1.60 \times 10^4}} = \sqrt{6.94 \times 10^{15}}$ $v = 8.33 \times 10^7\text{ m s}^{-1}$ (approx $0.28c$). (d) From Kepler's 3rd Law, $T^2 \propto r^3$. $\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3}$ Given $T_2 = 2T_1$, then $\frac{(2T_1)^2}{T_1^2} = 4$. $4 = \frac{r_2^3}{r_1^3} \implies r_2 = r_1 \times \sqrt[3]{4}$ $r_2 = (16.0\text{ km}) \times 1.5874 = 25.398\text{ km}$ Height $h_2 = r_2 - R = 25.398 - 12.0 = 13.398\text{ km}$ $h_2 = 13.4\text{ km}$ (to 3 SF).

Marking guidance

(a) - Correct calculation of volume $V = 7.24 \times 10^{12}\text{ m}^3$ [1] - Correct multiplication by density to reach $1.66 \times 10^{30}\text{ kg}$ [1] (b) - Correct substitution into $g = GM/R^2$ using $R = 1.2 \times 10^4\text{ m}$ [1] - Final answer $7.7 \times 10^{11}\text{ N kg}^{-1}$ (allow $7.71 \times 10^{11}$) [1] (c)(i) - Correct orbital radius $r = 1.6 \times 10^4\text{ m}$ [1] - Equating centripetal force to gravitational force or stating Kepler's 3rd Law [1] - Correct substitution of values into $T^2 = 4\pi^2 r^3 / GM$ [1] - Final answer $1.21 \times 10^{-3}\text{ s}$ (or $1.2\text{ ms}$) [1] (c)(ii) - Correct use of $v = \sqrt{GM/r}$ or $v = 2\pi r / T$ [1] - Final answer $8.3 \times 10^7\text{ m s}^{-1}$ [1] (d) - Use of $T^2 \propto r^3$ or $r^3/T^2 = \text{constant}$ [1] - Calculation of new radius $r_2 = r_1 \times 4^{1/3}$ or $25.4\text{ km}$ [1] - Subtracting star radius $R$ from orbital radius $r_2$ [1] - Final answer $13.4\text{ km}$ [1]