I’m curious if it would be similar to that of the moon, less than, or moreso.
It is made entirely of metal which is super dense and has a lot of mass, but a lot of it is also open space I imagine. Curious if the smaller amount of dense metal would have a larger or smaller mass than a solid moon.
Wait. I just realized energy also creates a gravitational pull, and the death star’s whole thing is destroying a planet right? That’s got to take a huuuge amount of energy because the explosion has to massively overcome the gravity holding the target together.
A quick google search says you’d need 10^32 Joules to blow up the earth. E=mc2 so dividing that energy by the speed of light squared gives about 1.1e15 kg of equivalent mass which is relatively small compared to earths mass (6e24) but still large.
For reference, if the radius of the Death Star was 1000m you’d get about 5.2m/s2 acceleration from just that energy in its core.
But if the Death Star is able to blow up multiple planets, then the energy it has to have on hand goes up. So if the Death Star contains enough energy to blow up 5.4 billion planets, then just that stored energy would have nearly equivalent “mass” to the earth.
But gravitational acceleration is inversely proportional to distance squared. So since the Death Star is small, you wouldn’t need that much energy to get earth gravity. If we assume the Death Star has about a 160km radius, then you’d only need enough stored energy to blow up ~45,000 earths to get a surface gravity of 9.1m/s2.
This gravity would increase as you got closer to the core or whatever part stores all that energy. But if you spread that energy out a bit you could probably extend how large the earth-like gravity range in the station would be.
The mass of the structure itself would contribute to the gravity too so that 45,000 is probably an overestimate.
TL;DR: From rough math in my head, assuming a radius of 160km, point mass, and ignoring the mass of the structure, you’d only need to store ~5e19 J of energy in the Death Star to get earth like gravity on the surface. That is approximately the amount of energy required to blow up 45,000 earths
As I understand it, gravity arises from the curvature of spacetime which is described by the Stress-Energy Tensor which only accounts for energy and momentum in a given part of spacetime.
So, really it’s like energy is actually what causes gravity in the first place, not mass. Massive things have large gravitational pull because mass itself has/is energy, E = mc2. This energy and its motion curves space and gravity results from that curvature.
Then again, I’m an engineering student not a physicist, so maybe I’ve got something wrong.
If it’s a big as a little moon it’ll have his own weak gravitational pull
It’s no moon
I’m curious if it would be similar to that of the moon, less than, or moreso.
It is made entirely of metal which is super dense and has a lot of mass, but a lot of it is also open space I imagine. Curious if the smaller amount of dense metal would have a larger or smaller mass than a solid moon.
Wait. I just realized energy also creates a gravitational pull, and the death star’s whole thing is destroying a planet right? That’s got to take a huuuge amount of energy because the explosion has to massively overcome the gravity holding the target together.
A quick google search says you’d need 10^32 Joules to blow up the earth. E=mc2 so dividing that energy by the speed of light squared gives about 1.1e15 kg of equivalent mass which is relatively small compared to earths mass (6e24) but still large.
For reference, if the radius of the Death Star was 1000m you’d get about 5.2m/s2 acceleration from just that energy in its core.
But if the Death Star is able to blow up multiple planets, then the energy it has to have on hand goes up. So if the Death Star contains enough energy to blow up 5.4 billion planets, then just that stored energy would have nearly equivalent “mass” to the earth.
But gravitational acceleration is inversely proportional to distance squared. So since the Death Star is small, you wouldn’t need that much energy to get earth gravity. If we assume the Death Star has about a 160km radius, then you’d only need enough stored energy to blow up ~45,000 earths to get a surface gravity of 9.1m/s2.
This gravity would increase as you got closer to the core or whatever part stores all that energy. But if you spread that energy out a bit you could probably extend how large the earth-like gravity range in the station would be.
The mass of the structure itself would contribute to the gravity too so that 45,000 is probably an overestimate.
TL;DR: From rough math in my head, assuming a radius of 160km, point mass, and ignoring the mass of the structure, you’d only need to store ~5e19 J of energy in the Death Star to get earth like gravity on the surface. That is approximately the amount of energy required to blow up 45,000 earths
Do you have a citation that energy has a gravitational pull?
General Relativity
As I understand it, gravity arises from the curvature of spacetime which is described by the Stress-Energy Tensor which only accounts for energy and momentum in a given part of spacetime.
So, really it’s like energy is actually what causes gravity in the first place, not mass. Massive things have large gravitational pull because mass itself has/is energy, E = mc2. This energy and its motion curves space and gravity results from that curvature.
Then again, I’m an engineering student not a physicist, so maybe I’ve got something wrong.
That’s pretty fascinating.