Solving a Gravitation Problem: Finding the Size of a Solid Gold Sphere

  • Thread starter Thread starter jakeowens
  • Start date Start date
  • Tags Tags
    Gravitation
AI Thread Summary
To determine the size of a solid gold sphere that achieves an acceleration due to gravity of 9.87 m/s², the relevant equations involve gravitational force and density. The density of gold is 19.3 x 10³ kg/m³, and the mass can be calculated using the formula M = (4/3)πr³ρ. By combining this with the gravitational equation g = GM/r², one can derive the necessary radius for the gold sphere. The discussion also touches on calculating the orbital period of the Apollo Lunar Modules, emphasizing the use of Kepler's Third Law without needing the mass of the module itself. Understanding these principles allows for solving both gravitational problems effectively.
jakeowens
Messages
34
Reaction score
0
Here's the problem I've been working on.

Gold has density of 19.3x103 kg/m3. How big would a solid gold sphere have to be if the acceleration due to gravity at its surface is to be 9.87 m/s2? (Check your answer against the radius of the Earth, which has a mean density of 5.5x103 kg/m3.)

Now I'm probably making this way to hard, and I am completely lost, and was wondering if anyone could help me out. The only way i could think to do this problem, was to calculate out the mass, then the radius, and volume of the earth. Then use the density of gold to find out how large a gold ball would have to be to be the same weight as earth.

But then when i get the radius of the gold ball and plug it into the equation g=G*(Me/Re^2) which should equal 9.87, but it equals like 15. so i know i screwed up somewhere.

I just can't think of how to do this problem.

any help is much appreciated
 
Physics news on Phys.org
Either my problems are incredibly hard, and expect way to much from you, or i am being stupid and trying to do things the hard way.

Each of the Apollo Lunar Modules was in a very low orbit around the Moon. Given a typical mass of 14.7 E 3 kg, assume an altitude of 64.0 km and determine the orbital period.

this problem has me stumped to. I find myself having to look up things on the internet, such as the mass of the moon and radius of the moon and crap like that. This can't possibly be that involved can it. So that's why i think I'm doing this all wrong, and was just wondering if anyone could point me in the right direction.
 
jakeowens said:
Gold has density of 19.3x103 kg/m3. How big would a solid gold sphere have to be if the acceleration due to gravity at its surface is to be 9.87 m/s2? (Check your answer against the radius of the Earth, which has a mean density of 5.5x103 kg/m3.)

Consider the equation for acceleration due to gravity:

g=\frac{GM}{r^2}

Consider also the equation relating mass, density, and size:

M=\frac{4}{3}\pi r^3 \rho

Can you see a way to combine those equations to get a size, given acceleration and density?

Forget about the earth, it's a red herring, all you need to know is its acceleration due to gravity (9.8 m/s^2).
 
jakeowens said:
Each of the Apollo Lunar Modules was in a very low orbit around the Moon. Given a typical mass of 14.7 E 3 kg, assume an altitude of 64.0 km and determine the orbital period.

Consider Kepler's Third Law:

P^2=\frac{4\pi^2a^3}{GM}

Hint: You don't need the mass of the module.
 
what does that p looking sign mean in your first post, is that density?
 
Alright i got the first problem now, thanks. Hadnt considered combining those 2 problems, going to work on the 2nd now :D
 
what do you mean i don't need the weight of the module? I'm having a hard time with this one. What values am i supposed to use for a? or M?
 
jakeowens said:
what do you mean i don't need the weight of the module? I'm having a hard time with this one. What values am i supposed to use for a? or M?

Kepler's Third Law relates the period of an orbiting body to the semimajor axis of its orbit and the mass of the body around which it orbits. In a circular orbit, the semimajor axis is just the radius of the orbit.
 
Back
Top