In my textbook, under the topic of gravitation, it states that if the centres of 2 planets, each of mass $M$ and separated by a distance $r$ and you have a point halfway between the centres of the planets, the gravitational field strength at that point is $0$.

I don't fully understand why that is. Is it because the point feels an equal force in each direction so the resultant gravitational force is $0$, resulting in $0$ gravitational field strength at that point?

Surely however gravitational field strength is a measure of how many Newtons of gravitational force a body feels per kg. In this case, shouldn't it be equal to $$2\times \frac{GM}{(0.5r)^2}=\frac{8GM}{r^2}$$ as it feels $\frac{GM}{(0.5r)^2}$ Newtons of force from 2 planets?

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    $\begingroup$ You are correct, it is because the two forces are in opposite directions and cancel each other. Remember Newton's second law $\endgroup$ – Wolphram jonny Nov 2 '20 at 21:37
  • $\begingroup$ @Wolphramjonny ok: $F=ma$. So there's no resultant force you mean? Does that mean my idea of gravitational field strength is faulty? How would you explan gravitational field strength? $\endgroup$ – A-Level Student Nov 2 '20 at 21:41
  • $\begingroup$ the net gravitational force strength is zero, because it does not add up, it cancels because they point in opposite directions. The gravitational field is the total gravitational force per unit mass, so it is also zero. $\endgroup$ – Wolphram jonny Nov 2 '20 at 22:00
  • $\begingroup$ @Wolphramjonny thanks for the clear explanation. Would you like to post that as an answer? $\endgroup$ – A-Level Student Nov 2 '20 at 22:21

You are correct. The two forces are in opposite directions and cancel each other.

Force is a vector quantity. When adding vectors the directions are as important as the magnitudes.

Perhaps you are confusing gravitational field strength $g=GM/r^2$ and gravitational potential $V=-GM/r$. The former is gravitational force per unit mass, so like force it is a vector. When adding field strength you use vector addition (eg the parallelogram rule). The latter is the work done in moving a unit mass from infinity to a point at distance $r$ from the mass $M$, so like work it is a scalar. When adding potentials due to several masses you do so algebraically, regardless of the direction of the mass which is creating the potential.

  • $\begingroup$ Thanks for the clarification. $\endgroup$ – A-Level Student Nov 15 '20 at 20:45

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