# How would I calculate the gravity on the surface of a large rectangular solid? [closed]

This story on Reddit involves a 1 km by 1 km by 1.79 lightyear block of lead in deep space, and I was wondering if it was at all feasible (though that question is probably out of scope). To ask a more answerable question, how would I determine the surface gravity of such an object? In particular, how much gravity would you experience if you were standing in the middle of one of the two square sides?

## closed as off-topic by Aaron Stevens, stafusa, Jon Custer, rob♦Aug 14 at 19:58

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• What is your definition of "how much gravity"? – Aaron Stevens Aug 13 at 19:50
• @AaronStevens "how much gravity" = how much acceleration due to gravity would you experience – Rob Watts Aug 13 at 20:10
• FWIW, lead's pretty soft, and gravity will tend to convert the block to a sphere, with a radius a shade over 15930 km, assuming I didn't mess up the arithmetic (I used Google Calculator). – PM 2Ring Aug 13 at 20:46
• @PM2Ring yes lead is soft, but consider a 1 mm by 1 mm version - all that put together into a sphere would give 0.005 m/s^2 surface gravity, but spread out over more than a lightyear? The gravitational pull seems like it would be negligible, allowing it to remain a rectangular prism. If you made it bigger at some point it would obviously pull itself together into a sphere, but where is that point? en.wikipedia.org/wiki/Hydrostatic_equilibrium#Planetary_geology suggests for spheres it's around 400 km for ice and somewhat more than that for rock. – Rob Watts Aug 13 at 22:25
• arxiv.org/abs/1206.3857 – G. Smith Aug 14 at 0:57