## Correct use of "is proportional to" symbol (alpha)

Hello, I am facing a problem that can be solved quite easily using the proportional symbol ( I think ), so I would like to try to use it! Only problem is.. I don't know exactly how to use it correctly...

The question is :An astronaut weighs 882N on Earth, determing the weight of the astronaut on Planet X, which has a mass 95.3 times that of Earth and a radius 8.9 times that of Earth.

So, $$g=\frac{Gm}{r^{2}}$$ and thus $$g\alpha\frac{m}{r^2}$$

So I wrote down
$$g\alpha\frac{m}{r^2}$$
$$g\alpha\frac{95.3}{79.21}$$

But of course this false... g is not proportional to 95.3/79.21.. lol. Can someone show me how to correctly show my work? Thanks. This would allow me to simply use this ratio to calculate his new weight.
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 Remember that G is the gravitational constant, ie. it always takes the value 6.67ishe-11 This constant turns the proportionality into an equality.

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Homework Help
 Quote by Checkfate Hello, I am facing a problem that can be solved quite easily using the proportional symbol ( I think ), so I would like to try to use it! Only problem is.. I don't know exactly how to use it correctly... The question is :An astronaut weighs 882N on Earth, determing the weight of the astronaut on Planet X, which has a mass 95.3 times that of Earth and a radius 8.9 times that of Earth. So, $$g=\frac{Gm}{r^{2}}$$ and thus $$g\alpha\frac{m}{r^2}$$ So I wrote down $$g\alpha\frac{m}{r^2}$$ $$g\alpha\frac{95.3}{79.21}$$ But of course this false... g is not proportional to 95.3/79.21.. lol. Can someone show me how to correctly show my work? Thanks. This would allow me to simply use this ratio to calculate his new weight.
Saying $g \propto m/r^2$ is equivalent to saying that $g = Gm/r^2$ where G is a constant (the proportionality constant) ie. g is a linear function of m and r2. If you want to perform mathematical operations you have to use the equality sign and the constant.

$$g_1 = \frac{GM_1}{r_1^2}$$

$$g_2 = \frac{GM_2}{r_2^2}$$

dividing, the constant falls out:

$$\frac{g_2}{g_1} = \frac{M_2}{M_1}\frac{r_1^2}{r_2^2}$$

AM

Recognitions:
Homework Help
 Quote by Checkfate Hello, I am facing a problem that can be solved quite easily using the proportional symbol ( I think ), so I would like to try to use it! Only problem is.. I don't know exactly how to use it correctly... The question is :An astronaut weighs 882N on Earth, determing the weight of the astronaut on Planet X, which has a mass 95.3 times that of Earth and a radius 8.9 times that of Earth. So, $$g=\frac{Gm}{r^{2}}$$ and thus $$g\alpha\frac{m}{r^2}$$ So I wrote down $$g\alpha\frac{m}{r^2}$$ $$g\alpha\frac{95.3}{79.21}$$ But of course this false... g is not proportional to 95.3/79.21.. lol. Can someone show me how to correctly show my work? Thanks. This would allow me to simply use this ratio to calculate his new weight.
g is proportional to $$\frac{m}{r^2}$$
 If you want to use the proportionality sign, then say $$g_{e}\propto \frac{M_e}{r_e^2}[/itex] and [tex]g_{x}\propto \frac{M_x}{r_x^2}[/itex] where $g_{e/x}$ refers to earth or planet x etc. Now you can say: [tex]\frac{g_x}{g_e}=\frac{M_xr_e^2}{r_x^2M_e^2}$$ $$g_x}=g_e\frac{M_xr_e^2}{r_x^2M_e^2}$$. By the way, the "proprtional to" symbol isn't alpha. In tex it's "\propto"... heres the difference: $$\alpha \ldots \propto$$ The first is alpha, the second is proptional to.