2019 A/L Physics Past Paper · Question 22 (English Medium)

Gravitational acceleration at the earth–moon midpoint

A celestial object is located at the mid point of the line joining the centres of the earth and the moon at a certain time. The mass of the moon is 0.01230.0123 times the mass of the earth. Assume that the distance between the centres of the moon and the earth is 6060 times the radius of the earth. The acceleration of the object due to the gravity of both the earth and the moon in terms of gg is approximately,

  1. (1) 1.1×10−6 g1.1 \times 10^{-6}\,g
  2. (2) 1.1×10−3 g1.1 \times 10^{-3}\,g
  3. (3) 3.3×10−2 g3.3 \times 10^{-2}\,g
  4. (4) 0.5 g0.5\,g
  5. (5) 1.0 g1.0\,g
Answer and explanation

Answer: (2) 1.1×10−3 g1.1 \times 10^{-3}\,g

What the question asks

We want:

The acceleration of the object because of both the earth and the moon.

The object is at the middle of the line between their centres, 30R30R from each.

Give it in terms of gg.

Relevant theory

At a distance rr from a ball-shaped mass MM the field is GMr2\dfrac{GM}{r^2}. It points to the mass.

On the earth\'s surface g=GMER2g = \dfrac{GM_E}{R^2}.

So at a distance rr the earth\'s field is:

g(Rr)2g\left(\dfrac{R}{r}\right)^2

The two fields add as vectors.

Hence the above intensity is a vector quantity directed towards the mass m.

Applying it to this question

There are two sources.

The earth: mass MEM_E, distance 30R30R.

The moon: mass 0.0123ME0.0123M_E, distance 30R30R.

The earth\'s field points to the earth.

The moon\'s field points to the moon.

At the middle these are opposite ways.

Reasoning and calculation

The earth\'s pull

gE=g900g_E = \frac{g}{900}

gE=GME(30R)2=g900g_E = \dfrac{GM_E}{(30R)^2} = \dfrac{g}{900}

It points to the earth.

The moon\'s pull

gM=0.0123 g900g_M = \frac{0.0123\,g}{900}

gM=G(0.0123ME)(30R)2=0.0123 g900g_M = \dfrac{G(0.0123M_E)}{(30R)^2} = \dfrac{0.0123\,g}{900}

It points to the moon.

gnet=g900(1−0.0123)g_{net} = \frac{g}{900}(1 - 0.0123)

1.1×10−31.1 \times 10^{-3}

gnet=0.9877900 gg_{net} = \frac{0.9877}{900}\,g

gnet≈1.1×10−3 gg_{net} \approx 1.1 \times 10^{-3}\,g

The result is about 1.1×10−3 g1.1 \times 10^{-3}\,g. It points to the earth.

The moon adds only a small change. It is 1.2 % of the earth\'s pull. It does not change the first two digits.

Reading the options

Look at the answers.

Only option (2) is 1.1×10−3 g1.1 \times 10^{-3}\,g.

So the answer is option (2).

Why the other options fail

30430^4 was used.

130\dfrac{1}{30}. The 30 was used only once.

It thinks the field falls in a straight line with distance.

This is the value on the surface.

Checks

(1−m)/(D/2)2(1 - m) / (D / 2)^2

Does it make sense?

If the moon had the same mass as the earth, the two pulls would be equal and opposite.

The result would be zero.

Our answer gives (1−1)=0(1 - 1) = 0 ✓.

Conclusion

gnet≈1.1×10−3 gg_{net} \approx 1.1 \times 10^{-3}\,g

gnet=g900(1−0.0123)≈1.1×10−3 gg_{net} = \dfrac{g}{900}(1 - 0.0123) \approx 1.1 \times 10^{-3}\,g

It points to the earth.

So the answer is option (2).

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