Tim Worstall

Private Space Flight

Little piece outlining the private efforts to get into space in The Guardian. All we need now is for this meme, this understanding of bureaucracies, to reach the editorial pages:

No one has left Earth orbit since the Apollo programme ended in 1972,
and during the intervening years of what one astronaut describes as
"sliding around" in shuttles and space stations, Nasa changed from
being a creative, risk-taking organisation to a bureaucratic poodle,
forever begging funds from reluctant politicians.
…
The one thing the organisation could cling to was its monopoly on space activity, which it guarded jealously.

That’s simply what happens, always, to all bureaucraceis. As there is no feedback process, it is inevitable.

This looks wrong to me though:

What often goes unappreciated is this: where entering earth orbit is difficult, requiring speeds of 17,500 mph,

That’s escape velocity, isn’t it? Not orbital?

In

6 responses

  1. Pete in Dunbar Avatar
    Pete in Dunbar

    No, it’s orbital velocity at around the height the shuttle operates. There a calculator here: http://liftoff.msfc.nasa.gov/academy/rocket_sci/orbmech/vel_calc.html
    Escape velocity is around 25,000mph. One of those things I remember from being a child during the Apollo missions…

    Reply
  2. Alan Peakall Avatar
    Alan Peakall

    Pete is quite correct, Tim; orbital velocity and escape velocity are in the ratio of 1 : sqrt(2).
    With R = radius of the Earth, Ve = escape velocity, Vo = orbital velocity, U(r) = gravitational potential, M = mass of the Earth and m = magnitude of a test mass, the law of conservation of energy and the radial component of the equation of motion are:
    0.5 m Ve ^ 2 = m (U(Infinity) – U(R)) = G M m / R
    m Vo ^ 2 / R = G M m / R ^ 2
    from which the result follows.
    This is A level physics, and thus apparently beyond the grasp of the Daily Telegraph’s science correspondent emeritus Adrian Berry, who, in his book “The Next Five Hundred Years”, garbles the algebra to the point of implicitly asserting that all astronomical bodies are the same density!
    Tim adds: Well, you certainly lost me with that maths but then I’m not a science correspondent, fortunately. Berry’s position owed something, I believe, to the fact that the family name of the people who owned The Telegraph was Berry.

    Reply
  3. James Avatar
    James

    In what way is going to the moon leaving Earth orbit? What do they think the moon is doing other than orbiting the Earth?

    Reply
  4. Neil Craig Avatar
    Neil Craig

    He means low Earth orbit.

    Reply
  5. gene berman Avatar
    gene berman

    Allan:
    Back in the day of the moon landings, what you call “A-level” physics was relatively common information. People aren’t any dumber now or even less informed; there have merely been some changes in the matters about which information gets publicized at common levels.

    Reply
  6. David Gillies Avatar
    David Gillies

    James, ‘leaving Earth orbit’ means moving from a region where the dominant contribution to the gravitiational forces acting on you comes from the Earth to a region where it does not (e.g the surface of the moon). It’s essentially equivalent to moving from one Hill sphere to another. Doing this from the earth does indeed need a lot of energy.
    Escape velocity is around 25,000 mph (11200 m/s). This corresponds to a specific kinetic energy of around 63 MJ/kg. If you can slowly climb up out the Earth’s gravitation well on a space elevator to, say, 150000km up, you get that for a fairly modest power input, but until then, it’s light the blue touch paper and retire to a safe distance.

    Reply

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