What is the force needed to destroy this superplanet?

[DeathstroketheHedgehog]

In this guy's oc lore, there is a planet that is about 10% the size of the obsevable universe. (obviously let's assume it doesn't collapse into a black hole)

Its contents are very similar to Earth, so you can scale Earth's characteristics and density to this size.

How much force is needed to destroy this planet in joules? Is this 3-A?

Other questions: In comparison to Earth, how large is its surface area, and what is the gravity at surface level?


Edit: Please provide calcs if you can.

 

[Yobo_Blue]

Sounds 3-B

 

[JustSomeWeirdo]

Probably 3-B

 

[DeathstroketheHedgehog]

Really? Thought it'd be 3-A, considering 1% of the universe is actually matter.

 

[Webcamparrot]

If it's 10% the size of universe then it would be easily 3-A because a large majority of the universe is empty. If it's only 10% the mass of the universe then it's mid-high 3-B.

Do you know the composition of the planet? I can try and make a calc for it if so.

 

[Promestein]

Destroying solid structures even as large as a star system tends to get 3-B results so something this big is very high-end 3-B or 3-A outright

 

[Saikou_The_Lewd_King]

I actually forgot about that. I remember that solar system sized tree that was 3-B just via destroying it.

I wonder what that means for beings the size of several stars.

 

[Antoniofer]

When you say size you mean radius or volume? Anyway:

• 10% of the volume means 0.4642 the radius of the observable universe: [(0.4642*4.4*10^26)/6371000]^5*2.24*10^32 = 7.5824*10^129 J

• 10% the radius of the observable universe: [(0.1*4.4*10^26)/6371000]^5*2.24*10^32 = 3.5194*10^126 J

 

[DeathstroketheHedgehog]

Sorry it took me a while, my laptop is acting up.

I mean 10% of the diameter of the observable universe.

Also, if it isn't too much, could you define the variables in the equation you used?

 

[Antoniofer]

All right:

• 4.4*10^26 m is the radius of the observable universe (according to wikipedia).
• 2.24*10^32 J is the GBE of the Earth (min value).
• The ^5 comes from the equation: K*r^5*¤ü^2 (GBE equation regarding density); K is a constant and ¤ü is Earth's density, so just used Earth's GBE as reference using the new value of r (observable universe's radius).

 

[DeathstroketheHedgehog]

I understand where you got your numbers and answer from now, thanks.