# Calculating weight on different planets

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I weigh 80kg here on earth.

Suppose I travel to a planet that is twice the size of earth, and twice the mass. Assuming an honest scale, what would I weigh on this planet? The knee-jerk response of 160kg is almost certainly wrong — that would make sense if the planet were earth-sized but still twice the mass.

But if we start from there and use Newton's (or is it Hookes' ?) inverse square law, then for a planet twice the size, twice the mass of earth wouldn't the scale read (½)² = ¼th of 160kg, or 40kg? Intuitively, that doesn't make much sense. I understand the universe is under no obligation to satisfy my intuition, but am I even right in my calculations?

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Use $g = \frac{-GM}{r^2}$ to calculate the surface gravity of the planet you care about. Then just compare that to Earth Norm (~$9.8 m/s^2$)

Edit: Had to remind myself how to do Latex. Been gone a while.

Edit 2: Looking at your numbers, the 40kg seems right, if I'm doing the math right (albeit, I'm doing it in my head, so I may be off by any number of orders of magnitude).

Edit 3: Actually, kg is a measure of mass, not weight, and your mass wouldn't change. Your weight (in Newtons) would change by 1/2.

On Earth, you weigh $80 kg \times 9.81 m/s^2 = 784.8 Newtons$.

On a planet twice the mass & twice the radius of Earth, you should weigh 392.4 Newtons.

Edited by Greg H.

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Thanks, Greg.

I'm beginning to think 40kg is right. We were given (twice the radius, twice the mass) as you might find for Kepler-11b in http://kepler.nasa.gov/Mission/discoveries/, so it's a real question.

OTOH, if you think about it, if a [earth-like] planet is twice the radius of earth it should have 8 times the mass of earth, leading to an attractive force of 8/4 what you would have on earth, so my 80kg on earth becomes 160kg on this other planet (not Kepler-11b).

So the reason I weigh only 40kg on Kepler-11b is that it's about 1/4th the density of earth. It's sort of a "popcorn planet".

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Twice the radius and the mass would mean that the planet's density decreases with size. If you take a similar composition and "compactness", hence density, then gravity increases like the radius.

Small asteroids and small moons are known to be fluffy, with density well below ice.

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As Enthalpy says, it's the density that matters. You can actually calculate surface gravity based on density using the following:

$g = \frac{4\pi}{3} Gpr$

Where $p$ is the density of the object at radius $r$.

Since $p = \frac{mass}{volume}$ you can compute the change in the surface gravity based on the changes in volume and mass of the planet as compared to earth.

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As Enthalpy says, it's the density that matters. You can actually calculate surface gravity based on density using the following:

$g = \frac{4\pi}{3} Gpr$

Where $p$ is the density of the object at radius $r$.

Since $p = \frac{mass}{volume}$ you can compute the change in the surface gravity based on the changes in volume and mass of the planet as compared to earth.

'Density of the object. . .' What is the effect of moving from the object's surface into an orbit? I assume we just apply the inverse square rule, right?.