<p>Saying <span class="math-container">$g\propto \frac{M}{R}$</span> is the same as saying that there exists a constant <span class="math-container">$k$</span> for which <span class="math-container">$g = k\frac{M}{R}$</span>. This constant <span class="math-container">$k$</span> is the same for every planet or moon.</p>
<p>In these equations, you can replace <span class="math-container">$M$</span> and <span class="math-container">$R$</span> by the mass and ratio of the Earth to get the proportionality for the Earth. Then you can replace <span class="math-container">$M$</span> and <span class="math-container">$R$</span> with the mass and ratio of any other moon or planet and get the proportionality for that object.</p>
<p>In the 2nd part, you must obtain</p>
<blockquote>
<p>the ratio between the gravitational acceleration of Earth and the gravitational acceleration of the Moon</p>
</blockquote>
<p>Here, you just have to write
<span class="math-container">$
\frac{g_E}{g_M}
$</span>
and replace <span class="math-container">$g_E$</span> and <span class="math-container">$g_M$</span> with the expressions you determined in the first question.</p>
<p><span class="math-container">$\frac{g_E}{g_M} = \frac{k\frac{M_E}{R_E}}{k\frac{M_M}{R_M}}$</span></p>
<p>As you can see, the <span class="math-container">$k$</span> cancels out in the numerator and the denominator and leaves you with an expression of <span class="math-container">$\frac{g_E}{g_M}$</span> as a function of <span class="math-container">$M_E$</span>, <span class="math-container">$M_M$</span>, <span class="math-container">$R_E$</span> and <span class="math-container">$R_M$</span>.</p>

CC BY-SA 4.0
2021-01-18T09:26:38.610Z
2021-01-18T09:26:38.610Z
usernumber
<p>The first part is just replacing <span class="math-container">$M$</span> with <span class="math-container">$M_E$</span> and so forth. You get two statements of proportionality, one with the <span class="math-container">$E$</span> values and one with the <span class="math-container">$M$</span> values.</p>
<p>The second part is algebraic division.  Mathematically if <span class="math-container">$a\propto  b$</span>,  and if <span class="math-container">$a_1,a_2,b_1,b_2$</span> are specific values that are in this proportion. <span class="math-container">$a_1=kb_1$</span> and <span class="math-container">$a_2=kb_2$</span> for some constant <span class="math-container">$k$</span>. Division gives <span class="math-container">$a_1/a_2 = b_1/b_2$</span>.  The constants  cancel</p>
<p>By division, you turn a statement of proportionally into an equation.</p>

CC BY-SA 4.0
2021-01-18T08:34:40.890Z
2021-01-18T08:34:40.890Z
James K
<p>This question is for an astronomy class of mine. I don't believe the question is particularly difficult, but rather I'm having trouble understanding what it wants me to do. We haven't covered ratios and proportionalities and as a result I'm very lost. Here's the question:</p>
<blockquote>
<p>The gravitational acceleration due to a planet or moon tells you how “heavy” you feel when you stand on the planet or moon.  Gravitational acceleration is proportional to the mass and radius of the planet or moon in the following way:
<span class="math-container">$g\propto \frac{M}{R^2}$</span></p>
<ol>
<li>Start by writing two proportionalities, one for the Earth, one for the Moon. Use the following variables:</li>
</ol>
<ul>
<li><span class="math-container">$M_{E}$</span> : mass of Earth</li>
<li><span class="math-container">$M_{M}$</span> : mass of Moon</li>
<li><span class="math-container">$R_{E}$</span> : radius of Earth</li>
<li><span class="math-container">$R_{M}$</span> : radius of Moon</li>
<li><span class="math-container">$g_{E}$</span> : gravitational acceleration of the Earth</li>
<li><span class="math-container">$g_{M}$</span> : gravitational acceleration of the Moon</li>
</ul>
<ol start="2">
<li>Following the example under the ratio section, divide the two proportionalities to form a ratio, i.e., the ratio between the gravitational acceleration of Earth and the gravitational acceleration of the Moon. In other words, find the following in equation form:
<span class="math-container">$$\frac{g_{E}}{g_{M}}$$</span></li>
</ol>
</blockquote>
<p>This isn't from a lack of effort. I'm just not entirely sure how to write or divide a proportionality.</p>

<astrophysics>
Earth and moon gravitational ratios and proportionalities
user37712
# Earth and moon gravitational ratios and proportionalities

(Asked by: username_0 on 2021-01-18)

This question is for an astronomy class of mine. I don't believe the question is particularly difficult, but rather I'm having trouble understanding what it wants me to do. We haven't covered ratios and proportionalities and as a result I'm very lost. Here's the question:

> The gravitational acceleration due to a planet or moon tells you how “heavy” you feel when you stand on the planet or moon. Gravitational acceleration is proportional to the mass and radius of the planet or moon in the following way:
> $g\propto \frac{M}{R^2}$
>
> 1. Start by writing two proportionalities, one for the Earth, one for the Moon. Use the following variables:
>
> * $M\_{E}$ : mass of Earth
> * $M\_{M}$ : mass of Moon
> * $R\_{E}$ : radius of Earth
> * $R\_{M}$ : radius of Moon
> * $g\_{E}$ : gravitational acceleration of the Earth
> * $g\_{M}$ : gravitational acceleration of the Moon
>
> 2. Following the example under the ratio section, divide the two proportionalities to form a ratio, i.e., the ratio between the gravitational acceleration of Earth and the gravitational acceleration of the Moon. In other words, find the following in equation form:
>    $$\frac{g\_{E}}{g\_{M}}$$

This isn't from a lack of effort. I'm just not entirely sure how to write or divide a proportionality.

**Tags:** astrophysics

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## 2 answers

### Answer by username_2 on 2021-01-18. Score: 6 | Accepted answer

The first part is just replacing $M$ with $M\_E$ and so forth. You get two statements of proportionality, one with the $E$ values and one with the $M$ values.

The second part is algebraic division. Mathematically if $a\propto b$, and if $a\_1,a\_2,b\_1,b\_2$ are specific values that are in this proportion. $a\_1=kb\_1$ and $a\_2=kb\_2$ for some constant $k$. Division gives $a\_1/a\_2 = b\_1/b\_2$. The constants cancel

By division, you turn a statement of proportionally into an equation.


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### Answer by username_1 on 2021-01-18. Score: 3

Saying $g\propto \frac{M}{R}$ is the same as saying that there exists a constant $k$ for which $g = k\frac{M}{R}$. This constant $k$ is the same for every planet or moon.

In these equations, you can replace $M$ and $R$ by the mass and ratio of the Earth to get the proportionality for the Earth. Then you can replace $M$ and $R$ with the mass and ratio of any other moon or planet and get the proportionality for that object.

In the 2nd part, you must obtain

> the ratio between the gravitational acceleration of Earth and the gravitational acceleration of the Moon

Here, you just have to write
$
\frac{g\_E}{g\_M}
$
and replace $g\_E$ and $g\_M$ with the expressions you determined in the first question.

$\frac{g\_E}{g\_M} = \frac{k\frac{M\_E}{R\_E}}{k\frac{M\_M}{R\_M}}$

As you can see, the $k$ cancels out in the numerator and the denominator and leaves you with an expression of $\frac{g\_E}{g\_M}$ as a function of $M\_E$, $M\_M$, $R\_E$ and $R\_M$.


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