Question

Falling Meteorite The velocity of a heavy meteorite entering the earth's atmosphere is inversely proportional to \(\sqrt{s}\) when it is \(s\) kilometers from the earth's center. Show that the meteorite's acceleration is inversely proportional to \(s^{2}\) .

Short answer

The acceleration of the meteorite is indeed inversely proportional to \( s^{2} \), as evident from the final acceleration equation \( a = -k'/(s^{2}\))

Step-by-step solution

Establish the Given Relation

We know that the velocity \(v\) of the meteorite is given as \(v = k(1/ \sqrt{s}) \) where \(k\) is some constant of proportionality and \(s\) is the distance from the earth's center.

Derive Velocity to Obtain Acceleration

We also know that acceleration \('a'\) is the rate of change of velocity, thus it is the derivative of velocity. To obtain the acceleration, we differentiate velocity with respect to \(s\): \(a=\frac{dv}{ds}\). The derivative of \(1/ \sqrt{s}\) is \(-1/2s^{3/2}\), hence \(a = k(-1/2s^{3/2}) = -k/(2s^{3/2}) \).

Simplify the Resulting Equation

We can simplify the equation to \(a = -k'/(s^{2}\)), where \(k'\) denotes \(k/2\), which is another constant of proportionality.