Two planets, A and B, may be treated as point masses. Planet A has mass 4.00×10^24 kg and planet B has mass 1.00×10^24 kg. The distance between their centres is 6.00×10^8 m. A null point is a point at which the resultant gravitational field strength due to A and B is zero. Take…

Answer: ((a)) the two field contributions must be equal in magnitude and opposite in direction; this can only happen between A and B, since outside that region both fields point the same way (toward whichever planet is nearer) and can never cancel ((b)) x = 4.00×10^8 m from the centre of A (i.e. 2.00×10^8 m from B) ((c)) gA = 1.67×10^-3 N kg^-1, and gB at the same point is also 1.67×10^-3 N kg^-1 but in the opposite direction, so the resultant is zero. Part (a): Condition for a null point At a null point the field strength due to A and the field strength due to B must be equal in magnitude but opposite in direction, so that their vector sum is zero. Between A and B, the field due to A po…

A Level Physics 9702: Null Point Between Two Gravitating Planets

Two planets, A and B, may be treated as point masses. Planet A has mass $4.00\times10^{24}\text{ kg}$ and planet B has mass $1.00\times10^{24}\text{ kg}$ . The distance between their centres is $6.00\times10^8\text{ m}$ . A null point is a point at which the resultant gravitational field strength due to A and B is zero.

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

(a) State the condition, in terms of the magnitudes and directions of the two individual gravitational field contributions, that must be satisfied at a null point. Explain why this point must lie on the line joining the centres of A and B, somewhere between them. [2]

(b) The null point lies on the line joining the centres of A and B, at a distance $x$ from the centre of A (so at a distance $(d-x)$ from the centre of B, where $d$ is the separation of A and B). Show that $\frac{x^2}{(d-x)^2} = \frac{M_A}{M_B}$ and hence calculate $x$ . [5]

(c) Calculate the magnitude of the gravitational field strength due to A alone at the null point, and use it to confirm that the resultant field strength there is indeed zero. [3]

Part (a): Condition for a null point

At a null point the field strength due to A and the field strength due to B must be equal in magnitude but opposite in direction, so that their vector sum is zero.

Between A and B, the field due to A points toward A (i.e. in one direction along the line), while the field due to B points toward B (the opposite direction), so the two contributions can oppose each other and potentially cancel. Outside the pair (beyond A, or beyond B), both fields point the same way, toward whichever planet is nearer, so they always add and can never cancel. The null point must therefore lie somewhere on the line joining the centres, between A and B.

Part (b): Locating the null point

At distance $x$ from A (and $(d-x)$ from B), the magnitudes of the two field contributions are: $g_A = \frac{GM_A}{x^2}, \qquad g_B = \frac{GM_B}{(d-x)^2}$

Setting these equal (from part (a)): $\frac{GM_A}{x^2} = \frac{GM_B}{(d-x)^2}$

$G$ cancels from both sides: $\frac{M_A}{x^2} = \frac{M_B}{(d-x)^2}$

Cross-multiplying and rearranging: $M_A(d-x)^2 = M_B x^2 \quad\Rightarrow\quad \frac{x^2}{(d-x)^2} = \frac{M_A}{M_B}$

as required.

Substituting $M_A=4.00\times10^{24}\text{ kg}$ and $M_B=1.00\times10^{24}\text{ kg}$ : $\frac{x^2}{(d-x)^2} = \frac{4.00\times10^{24}}{1.00\times10^{24}} = 4.00$

Taking the (positive) square root, since both $x$ and $(d-x)$ are positive distances: $\frac{x}{d-x} = \sqrt{4.00} = 2.00$

$x = 2.00(d-x) = 2.00d - 2.00x \quad\Rightarrow\quad 3.00x = 2.00d \quad\Rightarrow\quad x = \frac{2.00d}{3.00}$

With $d=6.00\times10^8\text{ m}$ : $x = \frac{2.00\times6.00\times10^8}{3.00} = 4.00\times10^8\text{ m}$

So the null point is $4.00\times10^8\text{ m}$ from A, and $d-x = 6.00\times10^8-4.00\times10^8=2.00\times10^8\text{ m}$ from B. Confirming it lies closer to the less massive planet B, consistent with part (a).

Part (c): Verifying the field strength at the null point

Field strength due to A alone, at $x=4.00\times10^8\text{ m}$ : $g_A = \frac{GM_A}{x^2} = \frac{(6.67\times10^{-11})(4.00\times10^{24})}{(4.00\times10^8)^2}$

Numerator: $6.67\times4.00=26.68$ , and $10^{-11}\times10^{24}=10^{13}$ , so the numerator is $2.668\times10^{14}$ .

Denominator: $(4.00\times10^8)^2=16.0\times10^{16}=1.60\times10^{17}$ .

$g_A = \frac{2.668\times10^{14}}{1.60\times10^{17}} = 1.6675\times10^{-3}\text{ N kg}^{-1} \approx 1.67\times10^{-3}\text{ N kg}^{-1}$

Check using B instead, at $(d-x)=2.00\times10^8\text{ m}$ : $g_B = \frac{GM_B}{(d-x)^2} = \frac{(6.67\times10^{-11})(1.00\times10^{24})}{(2.00\times10^8)^2} = \frac{6.67\times10^{13}}{4.00\times10^{16}} = 1.6675\times10^{-3}\text{ N kg}^{-1}$

The two magnitudes agree exactly ( $1.67\times10^{-3}\text{ N kg}^{-1}$ each), and since they point in opposite directions (toward A and toward B respectively), the resultant field strength at this point is indeed zero, confirming that $x=4.00\times10^8\text{ m}$ is the correct location of the null point.

Final answers

(a) The field magnitudes must be equal and the directions opposite; this can only occur between A and B, since outside the pair both fields point the same way.
(b) $x = \boxed{4.00\times10^8}\text{ m}$ from A (equivalently $2.00\times10^8\text{ m}$ from B)
(c) $g_A = g_B = \boxed{1.67\times10^{-3}}\text{ N kg}^{-1}$ , equal in magnitude and opposite in direction, so the resultant field at the null point is zero