1g (gravitational acceleration) Sphere of U238

• Scottingham
In summary, the conversation discusses the formula for surface gravity and how it relates to the mass and radius of a body. The participants also discuss the density of U238 and how to calculate the mass and radius of a sphere to reach 1g at its surface. They also mention the need for two equations and the potential for using graphing or algebra to solve the problem. Ultimately, they come to the conclusion that the required radius for a 1g planet made of U238 would be around 1800km.
Scottingham
How big would a sphere of U238 have to be to reach 1g at its surface?

Scottingham said:
How big would a sphere of U238 have to be to reach 1g at its surface?
Do you not know how to look up the density of U238? Do you not know how to calculate the mass of a sphere? Do you no know the equation for the force of gravity at the surface of a sphere? That is, what part of solving this problem do you not understand?

Physics is not my forte, yet.

If I understand correctly do I just solve for R (where g=1)?from http://nova.stanford.edu/projects/mod-x/ad-surfgrav.html:

You have learned that the surface gravity (g)of a body depends on themass (M)and theradius (r)of the given body.

The formula which relates these quantities is:g = G * M / r2

whereG is called the Gravitational constant.

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Scottingham said:
Physics is not my forte, yet.

If I understand correctly do I just solve for R (where g=1)?
right

You have learned that the surface gravity (g)of a body depends on themass (M)and theradius (r)of the given body.

The formula which relates these quantities is:g = G * M / r2

whereG is called the Gravitational constant.

OK, that's a start. Now you need to find the density of U238 and write an equation for the mass of a sphere of radius r.

Thank you.

So if I understand correctly, I'd need two equations then. One to figure out the mass (with the known density of U238) given a particular radius. And another to figure out the radius given a particular mass in order to equal 1g. I'm guessing I could use some method to graph the two and where they intersected in mass and radius would be my solution?

Scottingham said:
Thank you.

So if I understand correctly, I'd need two equations then. One to figure out the mass (with the known density of U238) given a particular radius. And another to figure out the radius given a particular mass in order to equal 1g. I'm guessing I could use some method to graph the two and where they intersected in mass and radius would be my solution?
Just solve algebraically. No need to graph.

Hm ... give that U238 is 3 or 4 times as dense as the Earth's average density, 5949 km seems a bit high, but I didn't do the math so maybe it's right. I'd double check it though. I don't like using those apps ... I prefer to do it myself. You should try it that way and see what you get.

Scottingham said:
Gets 5949 km which seems more reasonable given that the Earth's radius is 6,371 km
Hm. Back of napkin calcs suggest your number may be a little too high.
Density of Earth is about 5g/cm3
Density of Uranium is about 19g/cm3.
So, volume of U sphere should be about .26 (1 / 3.8) of Earth. (excluding any compression effects)
Then cube root of .26 (to go from volume to linear) is .64.
.64*6371 = 4082
I think that leads to a radius closer to 4100km.[EDIT: Dagnabit you! phinds!]

I was thinking that it wasn't quite small enough either, but could the cubic nature of the volume of a sphere lead to a slightly unintuitive answer?

@Dave I don't think it'd work that was as it wouldn't be a linear comparison.

Scottingham said:
@Dave I don't think it'd work that was as it wouldn't be a linear comparison.
I was busy fleshing out my answer. You can see where I took the cube root to convert volume ratio (.26) to radius ratio (.64).

That definitely seems reasonable. The gravity at the surface also has the inverse square law from the center of mass to consider as well though, which is why it is maybe bigger than it first would seem...or my equation is wrong.

DaveC426913 said:
[EDIT: Dagnabit you! phinds!]
Now, now. Don't whine. Being a day late and a dollar short is common for Canadians

Scottingham said:
That definitely seems reasonable. The gravity at the surface also has the inverse square law from the center of mass to consider as well though, which is why it is maybe bigger than it first would seem...or my equation is wrong.
You are correct. Surface gravity does not scale with the cube of radius. It scales linearly. The cube factor from volume and the square factor from the inverse square law cancel to leave a simple linear relationship of surface gravity to radius (neglecting compression effects).

If Uranium is 3.8 times denser than Earth on average then the required radius of a 1 g planet is a factor of 3.8 times less. Roughly 1500 km.

I think the right answer should be around 1800km... (at least that's what maths tells me)

What maths though? Where did the equations I found go wrong? And how does a cube factor and a square factor cancel each other out?

Scottingham said:
What maths though? Where did the equations I found go wrong? And how does a cube factor and a square factor cancel each other out?

Here is how I work these sorts of thing in my head:

Double the radius and you increase volume (and therefore mass) by a factor of 8. Double the radius and you reduce gravity for a given mass by a factor of 4. Factor of 8 divided by factor of 4 is a factor of 2. You've doubled radius and you've doubled surface gravity. Therefore it is a simple linear relationship.

The gravitational force an object of mass $m$ feels from another of mass $M$ in a distance $r$ is: $F= \frac{GMm}{r^2}$
The Newton's 2nd law for the object of mass $m$ is $F=ma$
Equating them:
$ma =\frac{GMm}{r^2}$
$a= \frac{GM}{r^2}$
Now the mass $M$ of a sphere of radius $r$ and density $\rho$ is the density*volume. $M= \rho V_{sphere} = \rho \times \frac{4}{3} \pi r^3$...

So the acceleration is:
$a= \frac{4 \pi G \rho r^3}{3 r^2}= \frac{4 \pi G \rho}{3} r$

So the distance $R$ at which $a=g$ is given by:
$R= \frac{3g}{4 \pi G \rho}$

For: $G= 6.674 \times 10^{-11} \frac{m^3}{s^2 kg}$ , $\rho= 19 \frac{g}{cm^3}=19000 \frac{kg}{m^3}$ and $g=10 \frac{m}{s^2}$:

$R\approx 1883 km$ and gets a little less if you take g=9.81 m/s2 and ρ=18900 kg/m3, around 1856 km

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@jbriggs444 I don't think that's right about gravity scaling linearly. Also, I think pi helps to rule out the linear relationships.

I took the surface gravity equation here:
http://nova.stanford.edu/projects/mod-x/ad-surfgrav.html -- g = G * M / r2

And substituted mass with the density of U238 (18.9 g/cm3) multiplied by the volume of a sphere, which is 4/3πr^3.

Since the radius was in km in the surface gravity equation on converted the density one which was in cm by multiplying by 100,000.

@ChrisVer That looks more correct that what I was trying to do. I think I got all flummoxed with the units. Thank you!

Scottingham said:
I think I got all flummoxed with the units
what units? I also made some edits.

Scottingham said:
@jbriggs444 I don't think that's right about gravity scaling linearly. Also, I think pi helps to rule out the linear relationships.
Pi has diddly squat to do with whether a relationship is linear. In this case the relationship is linear.

Very wrong, yes. Try this one.

"solve 9.8=6.67*10^-11(18900(pi*r^3*4/3))/r^2 for r"

Result: r=1.85588x106

I took the surface gravity equation here:
http://nova.stanford.edu/projects/mod-x/ad-surfgrav.html -- g = G * M / r2
That equation is fine.

And substituted mass with the density of U238 (18.9 g/cm3)
Or 18900 kg per cubic meter.

multiplied by the volume of a sphere, which is 4/3πr^3.
Helps if you leave the pi in the formula when you submit it to Wolfram.

Since the radius was in km in the surface gravity equation on converted the density one which was in cm by multiplying by 100,000.
That's fine for converting cubic centimeters to cubic meters. But still leaves you out by a factor of 1000 because grams are not kilograms.

Instead of solving for g=<that formula> you solved for 1=<that formula>. But in units of meters/sec2, g = 9.8, not 1.

1. What is a 1g Sphere of U238?

A 1g Sphere of U238 is a sphere of uranium-238 with a mass of 1 gram. This is equivalent to approximately 1.25 cubic centimeters in volume.

2. What is the significance of a 1g Sphere of U238?

The significance of a 1g Sphere of U238 lies in its use as a reference object for measuring gravitational acceleration. Due to its small size and uniform density, it experiences a constant acceleration of 9.8 meters per second squared (9.8 m/s²) when placed on the Earth's surface.

3. How is a 1g Sphere of U238 created?

A 1g Sphere of U238 is typically created by melting and casting a small amount of uranium-238 into a spherical mold. The resulting sphere is then polished and calibrated to ensure its accuracy as a reference object.

4. Can a 1g Sphere of U238 be used to measure gravitational acceleration on other planets?

Yes, a 1g Sphere of U238 can be used to measure gravitational acceleration on other planets, as long as their surface gravity is within a similar range to Earth's (e.g. Mars, Venus). However, the value of 9.8 m/s² may not be accurate on planets with significantly different surface gravities.

5. Are there any other materials that can be used as a reference for measuring gravitational acceleration?

Yes, there are other materials that can be used as a reference for measuring gravitational acceleration, such as a pendulum or a mass on a spring. However, a 1g Sphere of U238 is considered the most accurate and consistent method for measuring gravitational acceleration due to its small size and uniform density.

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