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Gravitational Potential Energy

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$$ U = mgh $$

Where

  • \(U\) is the gravitational potential energy of the object,
  • \(m\) is the mass of the object,
  • \(g\) is the acceleration due to gravity, and
  • \(h\) is the height of the object above the ground.

This is an approximation that can be made when the object is small compared to the distance between the object and the ground, and g is fairly constant in the domain of observation.

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Example #

Kevin throws a 400,000 kg spaceship straight up from the surface of the earth, with an initial velocity of 100 m/s. After some time, the spaceship reaches its peak height. Assuming no losses due to friction, what is this peak height?

Since no energy is lost due to air friction, we know that all of the kinetic energy gets converted to potential energy.

The initial kinetic energy is \(\scriptsize E_k = \frac{1}{2}mv^2 = 200,000 * 100^2 = 5.0 \cdot 10^{10} J\)

The final potential energy is \(\scriptsize 5.0 \cdot 10^{10} = mgh = 400,000 * 9.8 * h\)

So, we can solve for the height of the spaceship:

\(\scriptsize h = \frac{5.0 \cdot 10^{10}}{400,000 * 9.8} = 1.27 \cdot 10^4 m\)

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