Gravitational tractor for towing asteroids

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SUMMARY

A recent paper by astronauts Edward T. Lu and Stanley G. Love proposes the concept of using a spaceship as a "gravitational tractor" for towing asteroids. The thrust required to counteract gravitational attraction is mathematically represented, but discrepancies arise when calculating thrust values. Specifically, a 20-ton spacecraft can tow a 200m diameter asteroid with a density of 2g/cm³, requiring a thrust of 1 Newton. The calculations reveal inconsistencies, particularly regarding the mass of the asteroid and the thrust needed to maintain a hover position above it.

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tony873004
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Tex doesn't seem to work in preview mode anymore. So sorry for the hard-to-read formulas.

In a paper recently published astronauts Edward T. Lu and Stanley G. Love propose using a spaceship as a "gravitational tractor for towing asteroids".
http://arxiv.org/ftp/astro-ph/papers/0509/0509595.pdf

There's some math I don't understand in this paper:

The thrust required to balance the gravitational attraction is given by T cos[sin-1(r/d)+a] = GMm/d^2 = 1.12(p/2g/cm^3)(r/d)^3(m/2x10^4Kg)(d/100m). Thus a notional 20 ton spacecraft with a=20 degrees hovering one half radius above the surface (d/r=1.5) can tow a 200m diameter asteroid (r=100m) with density p=2g/cm^3 provided it can maintain a total thrust T=1N.

Since the formula is in the form x=y=z, I imagine that if I computed them separately, that I'd get 3 identical answers. I'll try the 2nd and 3rd approaches:

Since it gives the asteroid diameter as 200m, and its density as 2g/cc, this means the asteroid's mass (M) is 8377580410 kg. It states the spacecraft is 20 tons. Tons has a few different meanings. I'm assuming they mean metric tons or 20,000 kg. Distance from center of mass is 150m.

GMm/d^2
(6.67e-11 * 8377580409.57278 * 20000) / 150^2 = 0.496696224 N

1.12(p/2g/cm^3)(r/d)^3(m/2x10^4Kg)(d/100m)
1.12((2g)/2g/cm^3)(1.5)^3(20000Kg/2x10^4Kg)(150m/100m)
1.12(1)(1.5)^3(1)(1.5)
1.12*1.5^3*1.5 = 5.67 (no units. They all cancel, so how can this equal thrust? Also, there's no M (asteroid mass) in this formula. How can I compute the amount of thrust needed to hover over an asteroid without the asteroid's mass?)

0.4967 N does not equal 5.67 does not equal 1 N. What am I failing to understand?
 
Last edited:
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The spaceship is angling its thrusters away from the asteroid to avoid hitting it - that's why more thrust is needed than GmM/d^2.

Using google calculator, I also get .496 Newtons for the force.

The spaceship is angling it's thrusters roughly 60 degrees away from a straight-line thrust, which is why it needs 1 Newton of thrust rather than .5 Newton.

60 degrees comes from arcsin(1/1.5) which is close to 40 degrees, plus 20 more.

The second expression is dimensionless - but it should have a value of 1 with the parameters specified - except for the first constant, which is in Newtons.

The paper probably meant to write

Thus T = 1 N * (complicated dimensionless expression)

I don't know why they didn't write it that way :-(.
 

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