# NASA's DART impact changed asteroid's orbit forever in planetary defense test

#### steve_foston

"Give me a big enough lever and I will move the Earth" (Archimedes). Congratulations to the NASA DART team for making the solar system a little safer, after all we may have to consider moving the orbit of the Earth in 5bn years (or earlier) to remain habitable for life on this planet

Brenton Grimes

#### billslugg

Sun output increases 1% every 100 million years. Insolation varies by the inverse square of the distance. This means the distance from the Sun must be increased by 750,000 Km per 100 million years. That equates to 7 .5 meters per year.

Mass of Earth is 6e24 Kg, mass of Sun is 2e30 Kg..
Attractive force between Sun and Earth is = G * Ms * Me /r^2 or 3.5e18 Kg.
Moving the Earth 7.5 meters in one year against a force of 3.5e18 Kg requires 8e11 joules which is 8e11 watts.

The Earth probably can't be tugged on directly by a rocket ship. Anything on the surface would be negated by opposite force from atmosphere.

Best way is to capture a large asteroid and position it near the Earth but ahead of Earth in its orbit around the Sun. Its gravity would pull the Earth slowly and gradually speed it up to a higher orbit where it would settle down slower than it started. The year would become longer. The required watts is roughly equal to current US electrical production.

Pogo

#### Pogo

It'll be interesting to see if this changes the period of the solar orbit of the Didymos system significantly.

Oh, wait a minute! In the news conference video, the speaker said that it decreased the orbital period from 11h55m to 11h23m (faster by 32 minutes). Then the speaker said that NASA expected to 'slow' the orbit by about ten minutes, but it 'slowed' the orbit by 32 minutes. Huh?

#### Unclear Engineer

Yeah, Nelson messed-up when he said "slowed the orbit". What the impact did was slow the speed of the smaller asteroid by hitting it head-on as it was coming toward the impacting craft in its orbit around the bigger asteroid. Slowing the little rock made it drop into a lower orbit around the bigger rock, gaining speed as it descended. So, it ended-up going faster because it was slowed-down - orbital mechanics seem weird. And, in the lower, faster velocity orbit, it takes less time to get around the smaller orbital path, so its orbital period gets decreased.

Actually, the orbit becomes more elliptical. Assuming an initial circular orbit, the impact changed that so the orbit gets lower on the opposite side of the big asteroid, and returns to the same altitude as the impact location on the side where the impact occurred. (But, I don't know what the initial orbit was like - it is theoretically possible that the impactor could have hit at the lowest point of the orbit and changed an initially elliptical orbit into a more circular one.)

Regarding the orbit of the pair around the sun, I would expect that to change a bit, too, because momentum must be conserved. The effect should be something like what would have happened if the impactor had hit the combined mass of both the small and large asteroid at whatever angle to the orbital path it had when it hit. Without knowing that angle, I can't say whether it made the pair orbit the sun a tad faster or slower, and how it would have changed its orbit shape.

#### steve_foston

Sun output increases 1% every 100 million years. Insolation varies by the inverse square of the distance. This means the distance from the Sun must be increased by 750,000 Km per 100 million years. That equates to 7 .5 meters per year.

Mass of Earth is 6e24 Kg, mass of Sun is 2e30 Kg..
Attractive force between Sun and Earth is = G * Ms * Me /r^2 or 3.5e18 Kg.
Moving the Earth 7.5 meters in one year against a force of 3.5e18 Kg requires 8e11 joules which is 8e11 watts.

The Earth probably can't be tugged on directly by a rocket ship. Anything on the surface would be negated by opposite force from atmosphere.

Best way is to capture a large asteroid and position it near the Earth but ahead of Earth in its orbit around the Sun. Its gravity would pull the Earth slowly and gradually speed it up to a higher orbit where it would settle down slower than it started. The year would become longer. The required watts is roughly equal to current US electrical production.

Thanks for the explanation and the calculation you illustrate - it appears that Archimedes was right and you can move the Earth with a big enough lever. Something for future generations to consider as the Sun heats up in the future. Appreciate your input

billslugg

#### Brenton Grimes

Pogo and billslugg

#### billslugg

Welcome to Space dot Com!!!
News flash: "People are idiots. Film at 11".

DART destroyed on impact
LICIACube saw the impact but was going just as fast as DART, could not hang around to see the dust clear.
Must wait forESA's HERA observation craft that will go into orbit around them four years from now. HERA was supposed to be there waiting for DART but funding delays by the Europeans forced NASA to use a cubesat instead to watch the impact.

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Pogo

#### Unclear Engineer

Moving the Earth 7.5 meters in one year against a force of 3.5e18 Kg requires 8e11 joules which is 8e11 watts.

Wouldn't that be 8e11 joules / 1 year to get watts? That would make the power needed = 8e11 joules / (1 year x 365 days/year x 24 hours/day x 3600 seconds/hour) = 25,000 joules/second = 25,000 watts. Watts is power, which is energy per unit time.

That doesn't seem like a lot of power, since I have a 25,000 watt generator.

But, wait, there are problems with the 8x10^11 joules number, too.

Going back a bit further, Kg is mass, not force, so the force result should be in Kg-m/sec^2 which is "newtons" of force. Using Bill's numerical values for G, the mass of the Sun and Earth, and 93 million miles for the radius of Earths orbit, I get 3.62 x 10^22 Kg-m/sec^2. So, I think Bills force estimate should be 3.62 x10^22 Newtons of force. And then moving the Earth 7.5 meters against that force would take 7.5 x 3.62x10^22 = 2.72x10^23 newton-meters of energy. A newton-meter is a Joule of energy. So, my number is much higher than Bill's for the energy needed.

So, 2.72x10^23 Joules of energy over a year is still a continuous power level of 8.6x10^15 watts, which would be necessary to be applied with 100% efficiency to move the Earth that 7.5 meters farther away from the sun over a years time.

For comparison, global electric power generation in 2021 was 28.2 terawatt-hours = 2.82x10^13 watt-hours = 3.22x10^9 watt-years. So, that is about 3.22x10^9 / 8.6x10^15 = 0.000037 of the power needed to move the Earth 7.5 meters farther from the sun over a year.

And, that doesn't begin to address the question of how to apply that power to produce motion of the Earth. We would have to eject mass from the Earth at high velocity, which is far from 100% efficient if we are talking about launching weights to escape velocity.

#### billslugg

Yes, you are correct in that the force is in Newtons, not kilograms. I misstated watts as joules. Also committed several errors in math.

Let me try again:
Mass of Sun is 2e30 Kg
Mass of Earth is 6e24Kg
Distance between the two is 150 million km or 1.5e11 meters.
Gravitational constant is 6.6e-11 N⋅m2⋅kg^−2
US electrical generating capacity in 2021 was 1.2e12 watts

Attractive force between Earth and Sun is = G * Ms * Me / r^2
Multiply 6.6e-11 times 2e30 times 6e24 and divide by 1.5e11 squared
That is 3.5e22 newtons
Moving the Earth 7.5 meters against a force of 3.5e22 newtons requires
2.6e23 joules of energy

2.6e23 joules of energy divided by the number of seconds in a year (3.15e7) gives Watts.
Watts = 8.3e15
This is 6,900 times US electrical generating capacity.

Yes, a propulsion mechanism must eject something in the opposite direction in order to provide thrust. It can be matter or it can be light.

We might use a laser beam to propel the Earth. Only if it didn't heat up the atmosphere too much.
Laser wattage requirement = newtons times speed of light
3.5e22 newtons times 3e8m/s = 1e31 watts
This is 9e18 times US electrical generating capacity. Probably not an option.

We could put an asteroid ahead of the Earth and thrust it to move faster along our orbit but then the products of thrust would impinge on Earth, maybe not such a good idea.

Fortunately we have plenty of time to iron out these problems.

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#### Unclear Engineer

Bill, not sure what you are referring to with "Watts is joules per second, not joules per year." I think I have properly taken into account the number of seconds in a year, both in converting the power needed to move the Earth and in scaling the number of terawatts of electric power produced on Earth in 2021. But, we all make mistakes, especially in shifting decimal points in cosmic level calculations. So, let me know if you find an error in my math (or logic).

#### billslugg

Bill, not sure what you are referring to with "Watts is joules per second, not joules per year." I think I have properly taken into account the number of seconds in a year, both in converting the power needed to move the Earth and in scaling the number of terawatts of electric power produced on Earth in 2021. But, we all make mistakes, especially in shifting decimal points in cosmic level calculations. So, let me know if you find an error in my math (or logic).

What I am referring to is that one watt of power consumes one joule of energy per second. I have adjusted my calcs for number of seconds in a year.

A watt is not a joule / year it is a joule / second.

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