# [SPM Physics Objective Question] How to solve this question?

1. Rajah 3 menunjukkan J dan K ialah dua planet masing-masing berjisim 2 m dan 3 m. Kedua-dua objek dipisahkan dengan jarak X.
Diagram 3 shows J and K are two planets of masses 2 \mathrm{~m} and 3 \mathrm{~m}, respectively. Both separated by a distance of X.

Rajah 3 / Diagram 3
Sebuah roket berjisim m bergerak dari J ke K. Apakah jarak roket dari J apabila daya paduan . tarikan graviti adalah sifar?
A rocket with a mass of m travel from J to K. What is the rocketâ€™s distance from J when the resultant gravitational force on it is zero?

We solve this using Newtonâ€™s Law of Gravitation:

Assuming the rocket is located in a linear line between J and K, there will be TWO forces, acting on the rocket in opposite directions. Let force by J on the rocket be Fâ±Ľ, force by K on rocket be Fâ‚–.

Let distance between J and rocket be x, and distance between rocket and K be X-x.

Calculate force by J on rocket:

Fâ±Ľ = G(2m)(m)/xÂ˛ = 2GmÂ˛/xÂ˛

Calculate force by K on rocket (opposite direction):

Fâ‚– = -G(3m)(m)/(X-x)Â˛ = -3GmÂ˛/(X-x)Â˛

When the resultant gravitational force on it is zero, this implies the magnitude of Fâ±Ľ = Fâ‚–.

2GmÂ˛/xÂ˛ = 3GmÂ˛/(X-x)Â˛
2(XÂ˛ - 2Xx + xÂ˛) = 3xÂ˛
Remember, take X as a â€śknownâ€ť value, and x as the unknown, rearranging in terms of quadratic equation with respect to â€śxâ€ť,
xÂ˛ + 4Xx - 2XÂ˛ = 0

Where a = 1, b = 4X, a = 1 , c = -2XÂ˛

x = [-4X Â± sqrt(16XÂ˛ - 4(1)(-2XÂ˛))]/2
Taking only the positive root since distance between J and rocket must be positive (x > 0),
x = [-4X + sqrt(24X^2)]/ 2 = [-4X + 2sqrt(6) X]/2 = (sqrt(6) - 2) X = 0.45 X

Therefore, answer is A. 0.45 X

I must say, this is an interesting application of ADD MATHS in Physics.

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Wahh itâ€™s really an interesting explanation. Thank you very much