Charon is a moon of Pluto that has a mass equal to \( \frac{1}{9} \) that of Pluto - AQA - A-Level Physics - Question 12 - 2017 - Paper 2

Let the mass of Pluto be ( M ) and the mass of Charon be ( m = \frac{1}{9}M ).

The gravitational field due to Pluto at a distance ( x ) from its centre is given by:
$E_{Pluto} = \frac{GM}{x^2}$

The gravitational field due to Charon at a distance ( (d - x) ) from its centre (which is located ( d ) away from Pluto) is given by:
$E_{Charon} = \frac{Gm}{(d - x)^2} = \frac{G(\frac{1}{9}M)}{(d - x)^2}$

For the resultant field to be zero, we set these two expressions equal:
$\frac{GM}{x^2} = \frac{G(\frac{1}{9}M)}{(d - x)^2}$

Cancelling ( G ) and ( M ) from both sides gives:
$\frac{1}{x^2} = \frac{1}{9(d - x)^2}$

Cross-multiplying yields:
$9(d - x)^2 = x^2$

Expanding:
$9(d^2 - 2dx + x^2) = x^2 \implies 9d^2 - 18dx + 9x^2 = x^2$

Rearranging gives:
$8x^2 - 18dx + 9d^2 = 0$

Using the quadratic formula, ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ), where ( a = 8, b = -18d, c = 9d^2 ):
$x = \frac{18d \pm \sqrt{(-18d)^2 - 4(8)(9d^2)}}{2(8)} \implies x = \frac{18d \pm \sqrt{324d^2 - 288d^2}}{16} = \frac{18d \pm \sqrt{36d^2}}{16}$

Thus:
$x = \frac{18d \pm 6d}{16}$

This gives us two potential values:

1. ( x = \frac{24d}{16} = \frac{3}{2} d ) (not valid as it exceeds the distance to Charon)
2. ( x = \frac{12d}{16} = \frac{3}{4} d ) (valid)

Thus, the distance of X from Pluto is ( \frac{3}{4} d ).