Size of planet with 1/50 gravity of earth

  • Thread starter YummyFur
  • Start date
  • #1
97
0
Basically I'm wondering how big a rock in space would need to be to draw an average person to it with a gravitational force of a kilo or two. Say it was the average density of earth and it was spherical, what would its diameter be.
 

Answers and Replies

  • #2
Drakkith
Staff Emeritus
Science Advisor
20,978
4,793
I'm not sure I understand you. I believe mass is measured in kilograms, not force. How much gravitational force are you looking at getting?
 
  • #3
Ryan_m_b
Staff Emeritus
Science Advisor
5,844
711
Well I'm pretty sure you would need 50 times less mass. Every time you half the diameter of a sphere you reduce it's volume by 8. So a sphere 1/4 the size of Earth would have 1/65th of the gravity.
 
  • #4
1,973
264
Well I'm pretty sure you would need 50 times less mass. Every time you half the diameter of a sphere you reduce it's volume by 8. So a sphere 1/4 the size of Earth would have 1/65th of the gravity.
But you would als get closer to the center. Every time you halve the diameter, there will be 8 times less mass, the 1/r^2 in Newton's law. The end result is that an object with 1/4 times the diameter of Earth, will have 1/4 times the gravity, so you'll need 1/50 of the diameter.
 
  • #5
97
0
basically I was wondering if a future astronaut landing on an asteroid that just had enough gravity to keep him on the surface would be able to tell he was on a small object by it's appearance.

@ryan, that doesn't seem right because the moon is about 1/4 the diameter of the earth and is about 1/6th the gravity.

@drakkith, my bad. I meant how much would he weigh.
 
  • #6
97
0
@willem2, ok so then that would be say a rock that was about 160 miles across. So an astronaut standing in the middle would see about 80 miles to his horizon.
 
  • #8
DaveC426913
Gold Member
18,902
2,411
basically I was wondering if a future astronaut landing on an asteroid that just had enough gravity to keep him on the surface would be able to tell he was on a small object by it's appearance.
Yes. He would see a horizon that's very near. Mars' horizon is something like a couple of kilometers, enough to be quite noticeable.
 
  • #9
1,973
264
@willem2, ok so then that would be say a rock that was about 160 miles across. So an astronaut standing in the middle would see about 80 miles to his horizon.
The distance to the horizon on a perfect sphere is

[tex] \sqrt { 2 R h } [/tex]

where R is the radius of the sphere and h the height above the surface. This will be somewhat less than 1 km.
 
  • #10
26
0
The universal law of gravitation is:
[tex]F_g = \frac{mMG}{r^2}[/tex]
Where [tex]G = 6.674*10^{-11} \frac{m^3}{kg*s^2}[/tex]

The average mass density of the earth is 5.515 g/cm^3 (According to Wikipedia),
so the acceleration due to gravity based on a planet with Earth's density and radius r is [tex]a = \frac{\frac{4}{3}*\pi*r^3*5.515*10^6*6.674*10^{-11}}{r^2}[/tex]

Using gravity on Earth to be 9.807 m/s, we write the equation
[tex] 9.807/50 = 1.5418*10^{-6}*r[/tex]

This can easily be solved to find a radius of 127 kilometers.
 
  • #11
DaveC426913
Gold Member
18,902
2,411
There has got to be a formula that, for spherical objects of a fixed density, takes only mass and radius and returns the surface gravity.
 
  • #14
2,257
7
about 1000 km

ceres is the smallest one
 
  • #15
3,741
418
Check out this website:

By the way, what's the limit for mass and/or radius of a body to be gravitationally compacted into a sphere? (I used to know it, but I forgot.)
It should be dependent on the composition of the body. I doubt that there is a general answer.
 
  • #16
195
1
about 1000 km

ceres is the smallest one
About 1000km given what density?

The gravitational attraction of a planet or a moon is dependent on it's diameter and its density!
 
  • #17
195
1
It should be dependent on the composition of the body. I doubt that there is a general answer.
NO KIDDING!!!

See my last two posts on this thread.
 
  • #18
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,115
151
This can be solved without actually knowing the density of Earth, or plugging in the value of G. We just need to know the radius of the Earth.

The surface acceleration, for a sphere of mass M and radius r is
[tex]a = \frac{GM}{r^2}[/tex]
Since M = ρV, and V = 4πr 3/3, we have
[tex]a = \frac{4 \pi G \rho r^3}{3 r^2} = \frac{4 \pi G \rho}{3}r[/tex]
In other words, a and r are proportional for a fixed density ρ. To get 1/50th the force or acceleration as experienced at the surface of the Earth, the radius must be 1/50th that of Earth:
[tex]r = \frac{6378 \text{ km}}{50} = 128 \text{ km}[/tex]

This agrees with tmiddlet's calculation posted earlier, and with what others have been saying without showing a derivation.
 
  • #19
3,741
418
NO KIDDING!!!

See my last two posts on this thread.

Sorry, I cannot see any other posts between your claim that you used to know the answer and my remark. Still don't see why would you yell at me.
Your question:

"By the way, what's the limit for mass and/or radius of a body to be gravitationally compacted into a sphere? (I used to know it, but I forgot.)"

is not well defined as it depends on the initial state of the "body" to be (or not to be) compacted (whatever that means). I don't dispute that you may have had an answer but it should have been to a more specifically formulated question.
 
  • #20
Drakkith
Staff Emeritus
Science Advisor
20,978
4,793
There has got to be a formula that, for spherical objects of a fixed density, takes only mass and radius and returns the surface gravity.
Wouldn't it be the standard formula, but with the variable density replaced with just a number?
 
  • #21
DaveC426913
Gold Member
18,902
2,411
Wouldn't it be the standard formula, but with the variable density replaced with just a number?
Yes, 1. Had someone posted the formula? I did not see it. Or did not realize that was the general one.
 
  • #22
Redbelly98
Staff Emeritus
Science Advisor
Homework Helper
Insights Author
12,115
151
Yes, 1. Had someone posted the formula? I did not see it.
No, not before your original comment in Post #11.

But after it, in Post #18, I give a at the surface in terms of either (take your pick) (1) mass and radius or (2) average density and radius.
 
  • #23
195
1
nasu:

Sorry about the shouting. I should have put a smiley after that. I tai care to be more polite in the future.

And you're correct that the post to which you responded wasn't very clear, but when I used the phrase:

"...mass and/or radius of a body..."

I intended to refer to density, as the radius plugged into the sphere equation gives you the volume, and mass divided by volume gives density.
 
Last edited:
  • #24
97
0
I know it's probably buried in the formulas somewhere but I can't work this out... further to the original query, I am now wondering what the escape velocity would be for this 127km asteroid where the average person would weigh a kilo or two.

thx
 
  • #25
195
1
In other words, a and r are proportional for a fixed density ρ.
And for a different density, you could complete the equation by plugging in the absolute density, or by ignoring the gravitational constant and dividing your final result by the relative density. (In other words, if your density is 0.8 that of earth, your final answer would be 128km/0.8 = 160km.
 

Related Threads on Size of planet with 1/50 gravity of earth

  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
18
Views
4K
Replies
21
Views
4K
Replies
2
Views
451
  • Last Post
Replies
5
Views
2K
Replies
1
Views
3K
Replies
6
Views
2K
Replies
9
Views
993
Top