Expert: Philip Stahl - 3/2/2006

Question
Some object on Earth's surface has a weight of 81 kg. To translate that into mass I can use the equation m = w/g, where 'm' is mass, 'w' is weight, and 'g' is gravitational acceleration. For Earth this would become m = 81kg/9.8 m/sec^2 = 8.3

If this same object was on a planet with twice Earth's mass but the same radius, the value of 'g' would become 19.3 m/sec^2

Could you please show me how to determine the weight of the original object on this second planet?

Answer
Hello.

First of all, we need to examine the units. In the case of your object, 81 kilograms (kg) is the *mass*, not the weight. Weight is reckoned in Newtons or some derivative thereof, not kg.

Now, its weight would be: w = mg (analogous to F = ma, e.g. mass x acceleration. In this case, the acceleration of gravity).

So:  W  = (81 kg) (9.8 m/sec^2)  =  793.8  kg-m/sec^2


Now by dimensional and unit analysis we know that

1 N (Newton) =  1 kg-m/sec^2

Thus, the object in this case has a *weight* of 793.8 N

Again, let me emphasize that Newtons is a FORCE unit, and weight is a force! (Actually, the reaction force to Earth's downward pull at any given location)

In the case of the hypothetical planet you referenced - with g' = 19.3 m/sec^2, the mass is still the same at 81 kg.

The new weight, w', will be:

w' =   (81 kg) (19.3 m/sec^2)

w' =  1563.3 N

When you think about it, this makes sense, since a constant mass is now undergoing about twice the acceleration as before.

If one has:   w = mg

and            w' = mg'    (m = const)

one can arrive at the proportion:

w/ w'  =  g/ g'

so any new weight (w') would be expressed:  w' = (w g')/ g

Thus, if g' = 2g, then:

w' =    w (2 g)/g  =   2w

Hope this makes sense to you, and clarifies issues - the primary one of which is to always ensure you have consistency of units at the outset.