<p>For the most part, you can assume <a href="https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation" rel="nofollow noreferrer">Newton's law of universal gravitation</a>, (see link for more information). Basically, this law can be stated as the following.</p>

<p>For the gravitational force between two celestial objects (or really, any two objects with mass), the force is given by the equation,</p>

<p>$$ F_G = G\frac{m_1 m_2}{r^2} $$</p>

<p>Given the mass of the two objects, $m_1$, $m_2$, and the distance between them, $r$. The constant of proportionally given by $G$, the gravitational constant (not to be confused with $g$, the acceleration of an object with respect to the conditions on the surface on the Earth).</p>

<p>For the specific problem(s) that you are given, you can say that, using the information from Wikipedia, for the mass of the <a href="https://en.wikipedia.org/wiki/Sun" rel="nofollow noreferrer">Sun</a>, <a href="https://en.wikipedia.org/wiki/Earth" rel="nofollow noreferrer">Earth</a>, and <a href="https://en.wikipedia.org/wiki/Neptune" rel="nofollow noreferrer">Neptune</a>,</p>

<p>$$ m_\text{Sun} = 1.988 \times 10^{30} \text{ kg} $$
$$ m_\text{Earth} = 5.972 \times 10^{24} \text{ kg} $$
$$ m_\text{Neptune} = 1.024 \times 10^{26} \text{ kg} $$</p>

<p>Your question gives us a preliminary distance. The $r$ for the distance between the Sun and the Earth is given as $1 \text{ AU}$, the $r$ for the distance between the Sun and Neptune given as $30 \text{ AU}$. Note that $1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$. </p>

<p>Therefore, for the gravitational pull (really another way of saying the gravitational force) between the Sun and Neptune, </p>

<p>$$F_{N \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 1.024 \times 10^{26} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left( 5.983 \times 10^{12} \text{ m} \right)^2} $$</p>

<p>Therefore, for the gravitational pull between the Sun and the Earth, </p>

<p>$$F_{E \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 5.972 \times 10^{24} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left(1.496 \times 10^{11} \text{ m}\right)^2} $$</p>

<p>Solving for these quantities (note that the answer that one might get may be numerically different depending on the sources of the planatary constants),</p>

<p>$$F_{N \leftrightarrow S} = 3.796 \times 10^{20} \text{ N}$$</p>

<p>$$F_{E \leftrightarrow S} = 3.541 \times 10^{22} \text{ N}$$</p>

<p>A bit of extra information, the gravitational force felt by the Earth or Neptune from the Sun is exactly the same as the force that the Earth or Neptune acts upon the Sun, as per Newton's third law of motion. But, by now, I suspect that the problem is answered to where taking the ratio of the two is a trivial manner.</p>

<p>Suggested by uhoh:
The force exerted by the Sun on Neptune (and vice versa) is about 93.3 times weaker than that of the force exerted by the Sun on the Earth(and vice versa).</p>

CC BY-SA 3.0
2018-02-16T09:41:47.237Z
Nice work, but you're not done yet. The title asks for the *relationship* between the two quantities, not the actual values. The body of the question repeats it "How much weaker..." In fact, you didn't need the mass of the sun nor the value of $G$, see [my comment](https://astronomy.stackexchange.com/questions/25116/how-would-you-relate-the-respective-solar-gravitational-pulls-of-two-planets#comment42983_25116).
uhoh
CC BY-SA 3.0
2018-02-17T03:11:52.527Z
Yes, you do not need to know a lot of the information provided. In all actuality, only the masses of the planets and the distances are needed (the rest is taken care of when the ratio of the two is computed). However, I have assumed in my answer that taking a ratio of the computed values is a trivial manner that does not need elaboration on. If desired by the OP, I could put information as to the ratio explanation as you have suggested.
Sparrow
CC BY-SA 3.0
2018-02-17T13:52:33.650Z
So if you added "or about 93.3 times lower" then it would actually answer the question as it was asked.
uhoh
CC BY-SA 3.0
2018-02-24T02:08:40.007Z
Excellent! Thanks :-)  I can't up vote twice, hopefully someone else will!
uhoh
CC BY-SA 3.0
2018-02-16T09:17:02.730Z
2018-02-22T09:18:39.997Z
2018-02-22T09:18:39.997Z
Sparrow
<p>I was asked this question:</p>

<blockquote>
  <p>How much weaker is the gravitational pull of the Sun on Neptune at 30 AU, than it is on Earth at 1 AU?</p>
</blockquote>

<p>How would I even begin to determine this?</p>

CC BY-SA 3.0
2018-02-16T04:22:17.600Z
Newton's law of gravity.
StephenG - Help Ukraine
CC BY-SA 3.0
2018-02-16T05:29:14.730Z
https://en.wikipedia.org/wiki/Gravity#Newton's_theory_of_gravitation there is an equation. Hint: "How much weaker" means you take a ratio, so $G$ and $M_{SUN}$ will cancel.
uhoh
<planet><gravity><planetary-science>
How would you relate the respective solar gravitational pulls of two planets?
fredafreak
# How would you relate the respective solar gravitational pulls of two planets?

(Asked by: username_0 on 2018-02-16)

I was asked this question:

> How much weaker is the gravitational pull of the Sun on Neptune at 30 AU, than it is on Earth at 1 AU?

How would I even begin to determine this?

**Tags:** planet, gravity, planetary-science
> **Comments:**
> **username_1**: Newton's law of gravity.  
> **username_2**: https://en.wikipedia.org/wiki/Gravity#Newton's_theory_of_gravitation there is an equation. Hint: "How much weaker" means you take a ratio, so $G$ and $M_{SUN}$ will cancel.  

---

## 1 answers

### Answer by username_3 on 2018-02-16. Score: 3 | Accepted answer

For the most part, you can assume [Newton's law of universal gravitation](https://en.wikipedia.org/wiki/Newton%27s_law_of_universal_gravitation), (see link for more information). Basically, this law can be stated as the following.

For the gravitational force between two celestial objects (or really, any two objects with mass), the force is given by the equation,

$$ F\_G = G\frac{m\_1 m\_2}{r^2} $$

Given the mass of the two objects, $m\_1$, $m\_2$, and the distance between them, $r$. The constant of proportionally given by $G$, the gravitational constant (not to be confused with $g$, the acceleration of an object with respect to the conditions on the surface on the Earth).

For the specific problem(s) that you are given, you can say that, using the information from Wikipedia, for the mass of the [Sun](https://en.wikipedia.org/wiki/Sun), [Earth](https://en.wikipedia.org/wiki/Earth), and [Neptune](https://en.wikipedia.org/wiki/Neptune),

$$ m\_\text{Sun} = 1.988 \times 10^{30} \text{ kg} $$
$$ m\_\text{Earth} = 5.972 \times 10^{24} \text{ kg} $$
$$ m\_\text{Neptune} = 1.024 \times 10^{26} \text{ kg} $$

Your question gives us a preliminary distance. The $r$ for the distance between the Sun and the Earth is given as $1 \text{ AU}$, the $r$ for the distance between the Sun and Neptune given as $30 \text{ AU}$. Note that $1 \text{ AU} = 1.496 \times 10^{11} \text{ m}$.

Therefore, for the gravitational pull (really another way of saying the gravitational force) between the Sun and Neptune,

$$F\_{N \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 1.024 \times 10^{26} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left( 5.983 \times 10^{12} \text{ m} \right)^2} $$

Therefore, for the gravitational pull between the Sun and the Earth,

$$F\_{E \leftrightarrow S} = \left( 6.674×10^{-11} \text{ N m$^2$ kg$^{-2}$} \right) \frac{\left( 5.972 \times 10^{24} \text{ kg} \right) \left( 1.988 \times 10^{30} \text{ kg} \right)}{\left(1.496 \times 10^{11} \text{ m}\right)^2} $$

Solving for these quantities (note that the answer that one might get may be numerically different depending on the sources of the planatary constants),

$$F\_{N \leftrightarrow S} = 3.796 \times 10^{20} \text{ N}$$

$$F\_{E \leftrightarrow S} = 3.541 \times 10^{22} \text{ N}$$

A bit of extra information, the gravitational force felt by the Earth or Neptune from the Sun is exactly the same as the force that the Earth or Neptune acts upon the Sun, as per Newton's third law of motion. But, by now, I suspect that the problem is answered to where taking the ratio of the two is a trivial manner.

Suggested by uhoh:
The force exerted by the Sun on Neptune (and vice versa) is about 93.3 times weaker than that of the force exerted by the Sun on the Earth(and vice versa).

> **Comments:**
> **username_2**: Nice work, but you're not done yet. The title asks for the *relationship* between the two quantities, not the actual values. The body of the question repeats it "How much weaker..." In fact, you didn't need the mass of the sun nor the value of $G$, see [my comment](https://astronomy.stackexchange.com/questions/25116/how-would-you-relate-the-respective-solar-gravitational-pulls-of-two-planets#comment42983_25116).  
> **username_3**: Yes, you do not need to know a lot of the information provided. In all actuality, only the masses of the planets and the distances are needed (the rest is taken care of when the ratio of the two is computed). However, I have assumed in my answer that taking a ratio of the computed values is a trivial manner that does not need elaboration on. If desired by the OP, I could put information as to the ratio explanation as you have suggested.  
> **username_2**: So if you added "or about 93.3 times lower" then it would actually answer the question as it was asked.  
> **username_2**: Excellent! Thanks :-)  I can't up vote twice, hopefully someone else will!  

---

