A Death Star appears

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Wow! Good (or not so good) thing for Derrick Rose, a Death Star appears in the orbit of the Earth! This thing is the same mass as the moon, but it's much closer to the Earth. How close does it have to be in order to change Derrick's 40 inch + 1 mm jump into a 60 inch jump?

Well, first I need to find the gravitational constant that would do this. Luckily, I already have everything I need to solve for this. The vertical jumping velocity would be the same, and the jump height would be 60 inches, or 1.524 meters. All I need to do is plug it into this equation:



0=\left(4.462\frac{m}{s}\right)^2+2g\left(1.524m\right)      thus      g=-6.533333333333333\frac{m}{s^2}

That is a lot lower than the -9.8 m/s2 we've come to know and love.

To find the change in gravity constant the Death Star would have to cause, I harnessed the power of simple algebra.

 -9.7938\frac{m}{s^2}+g_{death\:star}=-6.533\frac{m}{s^2}     so     g_{death\:star}=3.260\frac{m}{s^2}

Now I can find the distance the Death Star would be from the earth's surface!

All I have to do is plug it back into the gravitational equation   g=\frac{Gm}{r^2}

Here goes nothin'    3.260\frac{m}{s^2}=\frac{\left(6.67e^{-11}m^3kg^{-1}s^{-2}\cdot7.348e^{22}kg\right)}{r^2}\:      aaaaaand    r=1,226,136.133m\:   or    r=1,226.136\:km

1,226 km is disturbingly close to the earth. Especially because of the fact that it is the Death Star. At least Derrick can jump 60 inches now!

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