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Imagine two bodies (very, very small compared to the size of the earth), 1(km) separated from each other (in the direction of the earth) at infinity. After they reach earth after an infinite long trip how far will they be separated due to the tidal force? Can you use the tidal force to calculate the distance?. This force which is the difference in gravitational attraction on the bodies, which varies with distance, is zero at infinity and very small if the first plate arrives on earth: g of the first body on arriving at earth (ignoring resistance and assuming the earth's radius $r=6,371*10^6(m)$:

$$g_{fb}=\frac{MG}{r^2},$$

where $g_{fb}$ is the acceleration of the first body on earth, $M$ the mass of the earth,$5,97219*10^{24}(kg)$ and G the universal gravitational constant, $6,67408*10^{-11} (m^3 kg^{-1} sec^{-2})$.

Filling in those values we find $g_{fb}=\frac{39,859*10^{13}}{40,489*10^{12}}=9,844(\frac m {sec^2}) $.

For the second body we, of course, don't know the distance to the earth because this has changed during the trip due to the tidal force (which is what we want to calculate). But let's assume it's just $1000(m)$ more, $r$ changes from $6,371$ to $6,370*10^6(m)$.

In this case

$$g_{sb}=\frac{39,859*10^{13}}{40,576*10^{12}}=9,823(\frac m {sec^{2}})$$

As you can see the difference is very small but actually bigger after the bodies made the trip from infinity.

My question: Can we make a calculation involving the (time and distance varying) tidal force acting on the two bodies the whole trip?

It's all just Newtonian mechanics in one dimension, so certainly other means are there for calculating the deviation from $1000(m)$ (if someone knows how, please don't hesitate to answer; I didn't dug into it so deep) when the first body hits the earth.

But my main question was if it could be done with the help of (tiny tiny tiny) tidal forces that influence the distance of the bodies.

P.S. Let's assume (dmckee was so kind to point that out to me; thanks for that; I wrongly assumed that this was clear, which it is of course not!) the initial velocity of the bodies is zero, they're radially alligned with earth, and that their mutual gravitation is zero, for the sake of simplicity. The last assumption is obviously not true (which plays a role at infinity, where the tidal force is zero), and you can even make a new question out of this: What is the ratio of the tidal force between two 1(kg) bodies, 1(km) apart, of which one is on the surface of the earth and the other radially above it? If it turns out that their mutual gravitation is bigger, we must reject this assumption. Only if it's much, much smaller we can make this assumption. But let's assume it for sake of the question.

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    $\begingroup$ You seem to have a variety of unstated assumptions here: that the initial separation is radial, that the test objects begin at rest relative the earth, that the test objects' mutual gravitation can be neglected relative the tidal effect due to the earth (which is non-trivial if you assume an infinite initial separation of the pair from the earth). $\endgroup$ – dmckee Nov 28 '17 at 0:52
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    $\begingroup$ Run the problem in reverse: launch, one after the other, two test masses on a radial parabolic trajectory (straight escape orbit). Their separation when the 2nd test mass is launched is the separation you're asking about in your problem. Now ask: what is their separation 'at infinity'? I have a hunch that their separation goes to zero for any finite initial separation. $\endgroup$ – Alfred Centauri Nov 28 '17 at 1:07
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Presumably, both objects are at rest at infinity, so when they reach Earth they are each traveling at escape velocity.

Courtesy of Alfred Centauri's suggestion, suppose we run the experiment in reverse. We launch the two objects from Earth at escape velocity (11.2 kps), the 2nd one 89ms (1/11.2 s) behind the first. The objects follow the same trajectory, so the 2nd object is always 89ms behind the first.

Initially, the separation is 1km, because the speed is so high. But as they get further from the Earth they get closer together, because the object further from the Earth is always traveling slightly slower than the object which is nearer. As they reach infinity, where they stop, their separation is infinitesimally small, because their speed is infinitesimally small.

What does this tell us? That for any finite initial separation the two objects reach infinity together. In order for them to be separated by 1km when they reach infinity, they would have to be launched an infinite time apart. So running the experiment with the objects falling from infinity, initially 1km apart, they would be an infinite time and distance apart when they reached Earth.

Alternatively, suppose we calculate the time $T$ which it takes for an object to free-fall to Earth from rest at a distance $R$. According to this answer $T$ is proportional to $R^{3/2}$. The increase $\Delta T$ in time for a small increase in distance $\Delta R$ is proportional to $R^{1/2}$. So as $R \to \infty$ then $\Delta T \to \infty$ also, even though $\Delta R$ remains finite. The objects arrive an infinite distance apart.

Conclusion: If the objects start at rest (or traveling slowly) at infinity, they will reach Earth an infinite time (and distance) apart.

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  • $\begingroup$ -1 What about the fact that the second launched mass, all the way to infinity, experiences a bigger (although almost not noticeable) force directed towards than its predecessor? This means that the lower speed of the first mass can be compensated by greater deceleration of the second launched body. You omitted the tidal force. And why should two bodies at infinity, occupying the same place in space not arrive at the same time on earth? What you actually say is that two objects without a separation at infinity can have any distance with respect to each other when arriving on earth from infinity $\endgroup$ – descheleschilder Nov 28 '17 at 17:32
  • $\begingroup$ Thank you for your feedback. 1. Why does the 2nd mass experience a bigger force? 2. Tidal force is gradient of gravitational force. It stretches objects but does not make them fall faster. 3. Yes : two objects which are infinitesimally close at infinity can arrive at Earth a large finite distance apart. When you deal with infinities that's the kind of result you get. $\endgroup$ – sammy gerbil Nov 28 '17 at 19:44

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