# Atomic Manhole in Orbit

**URL:** <https://boards.straightdope.com/t/atomic-manhole-in-orbit/429809>\
**Category:** Factual Questions\
**Created:** [December 13, 2007, 3:21am UTC](https://boards.straightdope.com/t/atomic-manhole-in-orbit/429809 "2007-12-13T03:21:45Z")\
**Posts on this page:** 1\
**Showing post:** 3

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**Author:** ![Bill\_Door](https://avatars.discourse-cdn.com/v4/letter/b/50afbb/32.png) [@Bill\_Door](https://boards.straightdope.com/u/Bill_Door)\
**Post date:** [December 13, 2007, 3:42am UTC](https://boards.straightdope.com/t/atomic-manhole-in-orbit/429809/3 "2007-12-13T03:42:26Z")

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Okay, let’s think about this. According to Wiki, low earth orbital velocity is around 7,800 m/s. Kinetic energy equals 1/2 the mass times the square of the velocity, so a 40 kilogram manhole cover at 7,800 m/s has 1.22\*10^9 Joules of kinetic energy. The specific heat of iron is 0.45 J/g-K. That’s enough energy to raise the temperature of 40 kg of iron 7,270 K. Iron boils at 3,134 K. I think we can put this one to bed.

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