Possibility of detection of Earth approaching objects by gravity?

[Sven E]

TL;DR

I want to share briefly some thoughts on the possibilities of detecting potentially dangerous objects which are on a colission track with Earth (Asteroids which might impact).

Hey everyone,

lately I read a german newspaper article about an Asteroid which passed Earth in a very small distance (asteroid 2020 QG, distance of closest approach was about 2950 km).
Here is another link about this event written in English: Article about 2020 QG (Businessinsider.com).

As far as I understand, one major problem with similar objects is that they originate from a direction pointing roughly to the Sun so that a detection by visible instruments and/or methods is very hard or even impossible.

Now to my question/idea regarding this topic:
Shouldn't it be possible to detect such objects (just) in time by analyzing some sort of gravity field changes?
What kind of measurement setup and what degree of accuracy can you think of to realize a detection apperetaure/network in order to get a reasonable detection rate of potentially dangerous objects like 2020 QG?

Or maybe the changes in the gravity field are just too small or detectable too late?
If you think the latter is the case I would be very grateful for some physics with numbers and formulas(*) to underline this argument.Sven

(*) Such as Newton's law of gravity (F = G * m*M / r^2)

 

[Ibix]

Asteroid densities seem to be around 2g/cm³. This one was about 10m across, so say the mass was around 2×10⁶kg. It was about 2×10⁶m away at closest approach, which makes the gravitational acceleration from it around 10^-18g at closest approach.

That's assuming you can look for gravitational force directly, which you can't. You have to look for tidal effects, which scale as $r^{-3}$, so subtract another six orders of magnitude (maybe slightly less, depending on your detector design). And note that this doesn't even give you early warning.

It would be much easier, I think, to put satellite telescopes in solar orbits so that they have a good angle on anything coming at us from our "blind spot".

 

[DaveE]

In addition, you would have to distinguish it from every other moving gravitational source: trucks, people, airplanes, dogs, other asteroids...

The LIGO prototype at Caltech was really good at detecting the traffic outside on California Blvd.

 

[tech99]

The Arecibo radio telescope can obtain radar images of objects such as this which optical telesopes cannot see. In addition, the Sun is not a particularly strong noise source for radio waves.

 

[Vanadium 50]

That's assuming you can look for gravitational force directly, which you can't.

Why not? It's time dependent. Your sensor would read 9.79999999999999999 m/s2 instead of 9.8.

It seems to me the bigger problem is noise (as mentioned upthread), Furtheremore, asteroids this size strike the Earth literally all the time. Depending on the size (revised estimates are more like 5m than 10m), it looks like this would have been a once-a-year impact. Not even a Chelyabinsk, much less a Tunguska - and nowhere near a dinosaur-killer.

 

[Ibix]

Why not?

I was thinking of space-based detectors precisely to avoid all the vibration issues of ground based ones. I agree that you could, in principle, detect gravitational force directly with ground-based detectors.

 

[DrStupid]

I was thinking of space-based detectors precisely to avoid all the vibration issues of ground based ones. I agree that you could, in principle, detect gravitational force directly with ground-based detectors.

Earth is not fixed in space. It is affected by gravity as well. All you can locally detect are the tidal forces. A ground-based detector would just give a higher reading because Earth is so huge and the tidal forces therefore much stronger compared to a satellite. But the basic principle is the same.

 

[Ibix]

All you can locally detect are the tidal forces.

Consider the extreme case of an identical Earth passing the Earth on a grazing trajectory. While its operators ducked, a weighing scale with a test mass on it would briefly register zero as the two planets passed, from symmetry. The same principle applies to unequal masses such as Earth and an asteroid.

The point is that the scale is not really measuring gravitational force - it's measuring the contact force between the planet and a test mass. That's a perfectly honest and detectable force. In a similar vein, you can't even really measure tidal force, but you can measure the force needed to keep two test masses at constant separation in a non-uniform field.

 

[DrStupid]

Consider the extreme case of an identical Earth passing the Earth on a grazing trajectory. While its operators ducked, a weighing scale would briefly register zero as the two planets passed, from symmetry.

That wouldn't work if the gravitational field of the second Earth would be homogeneous. The scale would't show any difference in that case. The reading of the scale is affected by the inhomogeneity of the additional field only. It doesn't change due to it's additional gravitational acceleration but due the difference between it's additional gravitational acceleration and the additional gravitational acceleration of Earth in the external field (aka tidal acceleration). This effect scales with 1/r³.

 

[Ibix]

That wouldn't work if the gravitational field of the second Earth would be homogeneous.

Just to be clear, you are saying that at the point half way between two identical masses, the Newtonian gravitational force is non-zero? If that is what you are saying, can you tell me which way it points?

 

[jbriggs444]

Just to be clear, you are saying that at the point half way between two identical masses, the Newtonian gravitational force is non-zero? If that is what you are saying, can you tell me which way it points?

We need to carefully examine the scenario and nail down the details.

Let us begin by hand-waving away pesky details about the fact that both planets are squishy and surrounded by atmospheres. We can pretend that they are rigid spheres in a vacuum. [Edit: Let's give them surface gravitational accelerations of 1.00 meters per second squared].

We have a spring scale sitting on the Earth 1 with a 1.00 kg mass on its pan. Naturally it reads 1.00 Newtons.

Earth 2 comes whizzing past on a grazing trajectory. At the point of closest approach, Earth 1 is subject to a net gravitational acceleration of 0.25 g -- its center of mass is 2 Earth radii from the center of Earth 2. Inverse square law, yadda yadda.

To account for this acceleration, the scale should now read 1.25 Newtons. But...

Earth 2 attracts the 1 kg mass with an upward force of 1.00 Newtons. So the scale's reading is reduced to 0.25 Newtons.

The 0.75 Newton discrepancy between the original reading and the modified reading is due to the difference between the tidal gravity of Earth 2 at a distance of 1 r versus 2 r.

While its operators ducked, a weighing scale with a test mass on it would briefly register zero as the two planets passed, from symmetry.

Symmetry here is lost between the scale on Earth 1 is accelerating at 0.25 g along with Earth 1 one way while the scale on Earth 2 is accelerating at 0.25 g along with Earth 2 the other.

A scale at rest at the barycenter would indeed show zero throughout the scenario. By symmetry.

 

[DrStupid]

Just to be clear, you are saying that at the point half way between two identical masses, the Newtonian gravitational force is non-zero?

No, I’m saying, that this is not what the scale reads. The reading does not result from the gravitational acceleration of the scale but from the difference between the gravitational accelerations of scale and Earth.

Just let's do the math:

If we have Earth with the mass $m_E$ at the position $r_E$, an asteroid with the mass $m_A$ at the position $r_A$ and a body with the mass $m_B$ on a scale that is located in the distance $\Delta r$ from the center of Earth than the total force acting on the body is the sum of the gravitational force from the asteroid, the gravitational force from Earth and the normal force from the scale:

$F_B = \gamma \cdot \frac{{m_B \cdot m_A \cdot \left( {r_A - r_B } \right)}}{{\left| {r_A - r_B } \right|^3 }} - \gamma \cdot \frac{{m_B \cdot m_E \cdot \Delta r}}{{\left| {\Delta r} \right|^3 }} + F_N = m_B \cdot a_B$

The force acting on Earth is composed of the gravitational force from the asteroid, the gravitational force from the body and the counter force of the normal force acting on the body (the scale is assumed to be part of Earth):

$F_E = \gamma \cdot \frac{{m_E \cdot m_A \cdot \left( {r_A - r_E } \right)}}{{\left| {r_A - r_E } \right|^3 }} + \gamma \cdot \frac{{m_E \cdot m_B \cdot \Delta r}}{{\left| {\Delta r} \right|^3 }} - F_N = m_E \cdot a_E$

If the scale is located in the Amundsen–Scott station than the body has the same acceleration as Earth

$a_B = a_E$

(Of course I could also use any other place on Earth, but I don't want to struggle with the rotation.) With the Distance $R = r_A - r_E$ between asteroid and Earth this results in

$F_N = \gamma \cdot m_A \cdot \frac{{m_E \cdot m_B }}{{m_E + m_B }} \cdot \left( {\frac{R}{{\left| R \right|^3 }} - \frac{{R - \Delta r}}{{\left| {R - \Delta r} \right|^3 }}} \right) + \gamma \cdot \frac{{m_E \cdot m_B \cdot \Delta r}}{{\left| {\Delta r} \right|^3 }}$

That is the reading of the scale. It is obviously not zero in your example above ($m_A = m_E$ and $R = 2\Delta r$).

But this is still not sufficient to detect the asteroid. In order to see if the asteroid is there or not, it needs to be compared with a reference. That is the reading without asteroid:

$F_0 = \gamma \cdot \frac{{m_E \cdot m_B \cdot \Delta r}}{{\left| {\Delta r} \right|^3 }}$

The impact of the asteroid on the reading of the scale is

$\Delta F = F_N - F_0 = - \gamma \cdot m_A \cdot \frac{{m_E \cdot m_B }}{{m_E + m_B }} \cdot \left( {\frac{{R - \Delta r}}{{\left| {R - \Delta r} \right|^3 }} - \frac{R}{{\left| R \right|^3 }}} \right)$

I already rearranged the term in the brackets so that you can easily see what that results in for $\left| {\Delta r} \right| < < \left| R \right|$:

$\Delta F \approx \frac{{\gamma \cdot m_A \cdot m_E \cdot m_B }}{{m_E + m_B }} \cdot \frac{d}{{dR}}\left( {\frac{R}{{\left| R \right|^3 }}} \right) \cdot \Delta r = \frac{{\gamma \cdot m_A \cdot m_E \cdot m_B }}{{m_E + m_B }} \cdot \left[ {\Delta r - 3 \cdot \frac{{R \cdot \left( {R \cdot \Delta r} \right)}}{{R^2 }}} \right] \cdot \frac{1}{{\left| R \right|^3 }}$The effect doesn't scale with 1/R² as the gravitational force but with 1/R³ as expected for a tidal force.

 

[Ibix]

Yes, you're correct. I neglected the acceleration of the Earth in my thinking.