The Big Mo, Force and Momentum

The Big Mo, Force and Momentum#

Example 3: back to the space station from florida.#

The Question: At the end of May, 2020, astronauts Douglas Hurley and Robert Behnken docked the SpaceX Crew Dragon Demo-2 (”Endeavour”) to the International Space Station. This is the first time that an American flight to the ISS has flown since 2011 when Russia became the sole vehicle. Here’s a picture from the ISS of the Endeavor docking.

examples/_images/mechanics/dragon.png

Let’s compare the Endeavor to our aircraft carrier, the USS Nimitz. Remember, it has:

  • Mass: 80 Million kg, \(80 \times 10^6\) kg.

  • top speed: 32 knots or 37 mph or 16.5 m/s

  • momentum (found in the previous example for \(v=10\)m/s of \(p=8 \times 10^8\) kg m/s

Nimitz

The Endeavor:

  • mass: \(10^4\) kg

  • momentum at docking: \(p_D=8 \times 10^7\) kg m/s

Notice that even though the Endeavor is smaller than one of the jets on the Nimitz deck and much less massive, its momentum is only a factor of 10 smaller than that of the Nimitz. How is that possible?

The Answer:

Of course the aircraft carrier was just crawling along at city-street speed. So the space capsule must be going very fast: how fast?

\[\begin{split} \begin{align*} p_E &= m_E v_E \\ v_E &= \frac{p_E}{m_E} = \frac{8 \times 10^7}{10^4} \\ v_E &= 8 \times 10^3 \text{ m/s} \end{align*} \end{split}\]

That’s the orbital speed of the ISS and of course the docked Endeavor. It’s moving right along at 18,000 mph.

Imagine the momentum of the ISS itself! Well, you don’t have to imagine it. You’ll calculate it next.

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