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In this question I asked about gravity and in the answers it came up that the magnitude is equal (of the gravity acting on the Sun and the of the gravity acting on the Earth)

Does magnitude simply mean it's strength?

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Couldn't this have been appended to that discussion? Or googled? – ZachMcDargh Nov 4 '10 at 5:14
@Zach It's a different question and should be in its own thread:… ; Generally "just Google it" responses are discouraged on SE sites. We want people to come here when they "just Google it"… – coneslayer Nov 12 '10 at 20:04
@coneslayer: +1. Thank you for the references to those questions on meta. – Robin Maben Nov 29 '10 at 14:45

3 Answers 3

up vote 1 down vote accepted

A force is a normal vector and a vector is characterized by its magnitude ( its norm, "lenght" in graphical representation ) and direction.

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"Does magnitude mean strength" saying a force has magnitude, doesn't really help. – Jonathan. Nov 3 '10 at 17:54
You already know the answer ...; I was saying that a force is just a normal vector, call it magnitude, strength, length, norm, size, whatever you like. The magnitude (strength) is not different from "vector" as Herb said. A vector has a direction and a magnitude (strength). – Cedric H. Nov 3 '10 at 17:58

Sure - a force has a magnitude (amount) and direction (vector).

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In the case of centripetal forces, like gravity, force is always directed towards the center of mass - it's a radial force. We can therefore study most of its properties by calculating its strength.

In particular, with gravity it is typical to use polar coordinates (e.g., in 2D, use angle and radius instead of $x$ and $y$). This leads to a gravitational force which only has a radial component. In this particular case the magnitude of the force vector is the same as the radial component, so it's really easy to calculate.

In the general case, the magnitude is the length of the force vector calculated by the square root of the dot product of the vector by itself. In classical coordinates $(x,y)$

$|F| = \sqrt{(F_x^2 + F_y^2)}$

where $F_x$ and $F_y$ are the $x$ and $y$ components of the force vector ${\bf F}$.

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