# How can I destroy earth with physics? [closed]

I want to destroy the whole earth using physics, I would like to learn some of the ways that can be used to achieve this. I tried using a nuclear bomb but it takes so long, and I can't wait that much: Why does it take so long to make a nuclear bomb?

What are some physics experiments and theories to help me destroy the earth? Note that I have all the money and privileges I need.

## closed as unclear what you're asking by Kyle Kanos, Brandon Enright, user10851, David Z♦Mar 8 '14 at 3:08

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• Send a space crew to divert an asteroid onto the earth. – Isopycnal Oscillation Mar 7 '14 at 20:00
• I don't think you need physics to do this. Politics is doing it just fine. – jerk_dadt Mar 7 '14 at 20:24
• You'll need a Death Star – Kyle Kanos Mar 7 '14 at 20:29
• If you can't wait that much maybe you don't have the willpower to destroy earth anyway – Claudiordgz Mar 7 '14 at 22:42
• Last December, I received a text message from the USGS which said there had been an earthquake near Polson, Montana with a magnitude of 22.0. My first thought was that the Solar System would have a new asteroid belt, but my calculations indicated that there probably wouldn't be enough left to form one; the energy released would be about 250,000 times Earth's gravitational binding energy, more than enough to vaporize the planet. Unfortunately for your evil schemes, a followup message revised the magnitude to 2.2. – Keith Thompson Mar 8 '14 at 2:41

Earth's gravitational binding energy is $-1.711×10^{32}~\mathrm{J}$ or $4.09×10^{13}$ gigatons. The Tsar Bomba massed $27$ tonnes to deliver $0.057$ gigatons. Do the math for Earth disassembly by bomb. Substituting depleted uranium for the used lead tamper will double the yield.
Earth orbital speed averages 30 km/s and it masses $5.97×10^{24}~\mathrm{kg}$, so $mv^2 / 2 = 2.7×10^{33}~\mathrm{J}$. Rather than going for a messy asteroid impact and endangering Venusian hortas with debris infall, drop the Earth into the Sun, $t = 64.56$ days.
Patience. The Earth's rotational energy is about $2.13 \times 10^{29}~\mathrm{J}$ (non-homogeneous sphere). You'll need obtain the orbital stopper energy elsewhere.