Coming down to Earth

Problem: an object initially at rest some distance above the Earth begins to fall toward the Earth. Show that the time it takes for the object to fall halfway to the Earth is about 9/11 of the time it takes to fall all the way to the Earth.

Solution: use conservation of energy. Let \(R\) be the initial distance of the object above the Earth. Then the total energy of the object in its initial state is its gravitational potential energy:

\(– \frac{G M_e m}{R}\).

Note that the factors in the numerator correspond to the Universal Gravitation constant, the mass of the Earth, and the mass of the object, respectively.

At some point \(r\) between the initial distance and the Earth, the object will have a different value of the potential energy, as well as a nonzero value of kinetic energy. The statement of conservation of energy becomes

\(\frac{1}{2} m \dot{r}^2 – \frac{G M_e m}{r} = – \frac{G M_e m}{R}\).

The time \(t_0\) the object takes in falling to some position \(r_0\) is then found by solving this differential equation:

\(t_0 = \frac{1}{\sqrt{2 G M_e}} \displaystyle \int_{r_0}^{R} dr \left ( \frac{1}{r} – \frac{1}{R} \right )^{- \frac{1}{2} } = \sqrt{\frac{R^3}{2 G M_e}} \displaystyle \int_{x}^{1} du \left ( \frac{1}{u} – 1 \right )^{- \frac{1}{2} }\),

where \(x = r_0/R\). The integral on the right hand side may be evaluated through an appropriate substitution (details left to the reader):

\(\displaystyle \int_{x}^{1} du \left ( \frac{1}{u} – 1 \right )^{- \frac{1}{2} } = \arccos{ \sqrt{x} } + \sqrt{x(1-x)}\),

where the branch of the inverse cosine is the principal branch.

The fraction of time it takes for the object to fall halfway to Earth is the ratio \(p\) of the above expression at \(x = 1/2\) to the same at \(x = 0\):

\(p = \frac{1}{2} + \frac{1}{\pi} \approx 0.818\),

which is about 9/11.

 

Be Sociable, Share!

Published by

Ron Gordon

Math nerd in his early 40's who seems to have an opinion about everything and an inability to keep it to himself.