A Level Physics 9702: Newton's Law of Gravitation and Field Strength

The dwarf planet Kerrigan has mass $1.80\times10^{21}\text{ kg}$ . Kerrigan is a uniform sphere, so for any point outside it, its mass may be treated as a point mass located at its centre.

Take $G = 6.67\times10^{-11}\text{ N m}^2\text{ kg}^{-2}$ .

(a) State Newton's law of gravitation for the force between two point masses, giving it as an equation and defining each symbol used. [2]

(b) Calculate the gravitational field strength at a point $2.50\times10^6\text{ m}$ from the centre of Kerrigan. [2]

(c) A space probe of mass $420\text{ kg}$ is at the point described in (b). Calculate the magnitude of the gravitational force acting on the probe. [2]

(d) An identical probe is instead placed at a point $5.00\times10^6\text{ m}$ from the centre of Kerrigan, twice the distance used in (b). By considering how gravitational field strength depends on distance from a point mass, determine the new field strength at this point without recalculating $GM/r^2$ from scratch. [2]

Part (a): Newton’s law of gravitation

Newton’s law of gravitation states that any two point masses attract each other with a force proportional to the product of their masses and inversely proportional to the square of the distance between them: $F = \frac{Gm_1m_2}{r^2}$

where $F$ is the magnitude of the (equal and opposite) gravitational force on each mass, $G$ is the gravitational constant, $m_1$ and $m_2$ are the two point masses, and $r$ is the distance between their centres.

Part (b): Gravitational field strength at $2.50\times10^6\text{ m}$

Combining $F=GMm/r^2$ with $g=F/m$ for a point mass gives $g=GM/r^2$ . With $M=1.80\times10^{21}\text{ kg}$ and $r=2.50\times10^6\text{ m}$ : $g = \frac{GM}{r^2} = \frac{(6.67\times10^{-11})(1.80\times10^{21})}{(2.50\times10^6)^2}$

Numerator: $6.67\times1.80=12.006$ , and $10^{-11}\times10^{21}=10^{10}$ , so the numerator is $1.2006\times10^{11}$ .

Denominator: $(2.50\times10^6)^2 = 6.25\times10^{12}$ .

$g = \frac{1.2006\times10^{11}}{6.25\times10^{12}} = 0.019210\text{ N kg}^{-1} \approx 1.92\times10^{-2}\text{ N kg}^{-1}$

Check by recomputing differently: $1.2006/6.25=0.19210$ , and $10^{11}/10^{12}=10^{-1}$ , giving $g=0.19210\times10^{-1}=0.019210\text{ N kg}^{-1}$ . Both routes agree.

Part (c): Gravitational force on the probe

Using $F=mg$ with the field strength found in (b) and probe mass $m=420\text{ kg}$ : $F = mg = 420\times0.019210 = 8.068\text{ N} \approx 8.07\text{ N}$

Check using Newton’s law directly: $F=\dfrac{GMm}{r^2}=\dfrac{(6.67\times10^{-11})(1.80\times10^{21})(420)}{6.25\times10^{12}}=\dfrac{1.2006\times10^{11}\times420}{6.25\times10^{12}}=\dfrac{5.0425\times10^{13}}{6.25\times10^{12}}=8.068\text{ N}$ . Both methods agree, so $F\approx8.07\text{ N}$ .

Part (d): Field strength at twice the distance

Since $g=GM/r^2$ , the field strength is inversely proportional to the square of the distance ( $g\propto1/r^2$ ), not to the distance itself. Doubling $r$ therefore reduces $g$ by a factor of $2^2=4$ : $g_{\text{new}} = \frac{g}{4} = \frac{0.019210}{4} = 0.0048024\text{ N kg}^{-1} \approx 4.80\times10^{-3}\text{ N kg}^{-1}$

Check by recalculating directly: $g_{\text{new}}=\dfrac{GM}{(5.00\times10^6)^2}=\dfrac{1.2006\times10^{11}}{2.50\times10^{13}}=0.0048024\text{ N kg}^{-1}$ , which matches the ratio method exactly.

Final answers

(a) $F=\boxed{\dfrac{Gm_1m_2}{r^2}}$ , with $G$ the gravitational constant, $m_1,m_2$ the two point masses, and $r$ the distance between their centres
(b) $g=\boxed{1.92\times10^{-2}}\text{ N kg}^{-1}$
(c) $F=\boxed{8.07}\text{ N}$
(d) $g_{\text{new}}=\boxed{4.80\times10^{-3}}\text{ N kg}^{-1}$ (one quarter of the value in (b), since distance doubled and $g\propto1/r^2$ )