Correct use of is proportional to symbol (alpha)

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Homework Help Overview

The discussion revolves around the correct use of the "is proportional to" symbol in the context of gravitational force calculations. The original poster presents a problem involving an astronaut's weight on Earth compared to a hypothetical Planet X, which has different mass and radius values.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to express gravitational acceleration as proportional to mass and radius, but questions the validity of their approach when substituting values. Some participants clarify the relationship between proportionality and equality in gravitational equations.

Discussion Status

Participants are exploring the correct application of the proportionality symbol in gravitational equations. Some guidance has been offered regarding the distinction between proportionality and equality, as well as how to properly set up the equations for comparison between Earth and Planet X.

Contextual Notes

There is a focus on ensuring clarity in the use of symbols, particularly distinguishing between the proportionality symbol and the Greek letter alpha. The original poster expresses confusion about how to properly show their work in the context of the problem.

Checkfate
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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.
 
Last edited:
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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.
 
Checkfate said:
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
 
Checkfate said:
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}
but when cramming in the values you did, you get a comparison to Earth's "g." Multiply (95.3/79.21) by 9.81, and you get the "g" of the other planet.
 
If you want to use the proportionality sign, then say

g_{e}\propto \frac{M_e}{r_e^2}[/itex]<br /> <br /> and<br /> <br /> g_{x}\propto \frac{M_x}{r_x^2}[/itex]&lt;br /&gt; &lt;br /&gt; where g_{e/x} refers to Earth or planet x etc. Now you can say: &lt;br /&gt; &lt;br /&gt; \frac{g_x}{g_e}=\frac{M_xr_e^2}{r_x^2M_e^2}&lt;br /&gt; g_x}=g_e\frac{M_xr_e^2}{r_x^2M_e^2}.By the way, the &amp;quot;proprtional to&amp;quot; symbol isn&amp;#039;t alpha. In tex it&amp;#039;s &amp;quot;\propto&amp;quot;... here&amp;#039;s the difference:&lt;br /&gt; &lt;br /&gt; \alpha \ldots \propto&lt;br /&gt; &lt;br /&gt; The first is alpha, the second is proptional to.
 
Last edited:
Thanks a lot guys! :)
 

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