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If you draw the following diagram: you can see that the volume of the crater is the volume of the "truncated inverted cone" plus the volume of the bit of sphere. Since the volume of a cone is $V=\frac13 A h$ where $A$ is the area of the base and $h$ is the height, the volume of the truncated cone is given by $$V_{cone} = ...


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Couple of things to point out here, and I'm not sure whether either of those answer your question, so here are some ideas to help you think about the problem and guide you towards the full answer: The lowest possible value of Earth's gravitational potential is met: a)if the object is at Earth's exact center (r=0, namely there is no mass "below" him (m1=0), ...


1

The equation you cited makes use of "big G", the universal gravitational constant. Generally, this equation is used if you want to calculate the attractive force between two bodies, such the moon and the Earth, or a satellite and the Earth, or the Earth and the Sun, or if you want to calculate escape velocity from he Earth's gravitational field. The center ...


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We treat a mass producing gravity as if all the mass is concentrated at the centre. Does that make sense to you? Also potential energy values are relative, it is only the difference in two values of P.E. that counts, not any absolute value. But anything placed on the surface of earth will never fall any deeper into the earth. We can assign a value of ...


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You could use two approaches. Sounds like now you're trying to carefully characterize and measure the shape of the crater, and then calculate the displayed volume. It sounds like the shape isn't completely regular, though, so you may have a tough time doing this accurately. Another approach would be to empirically measure the displaced volume of sand. Let's ...


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The Wikipedia article refers to a fireball, but as Wikipedia itself explains the word fireball has many meanings and doesn't necessarily literally mean a fire as in the combustion of a material in oxygen. In this case it means a ball of very high temperature gas. The gas is heated by the impact and gets hot enough to emit light just like the gas heated in a ...


5

The issue is that there are multiple crater scaling laws, each with different assumptions, as Horedt & Neukem (1984) show (title of paper is Comparison of six crater-scaling laws). Your first equation is a naive approach that the volume of the hole is linearly related to the kinetic energy, hence the $d= k\cdot E^{1/3}$ relationship. This particular ...


2

They say that gravity decreases as we dig into the earth. That's an immediate consequence of an overly simplistic model of the Earth, that the Earth is of a uniform density throughout. This is very far from the case. But I also read that gravity increases for the first approx. 2000km of distance underground Actually, it's about 2900 km ...


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The local gravitational field increases (slightly) as you descend under the earth because much of the earths mass is in the core, and you're getting closer to the core. As the moon has a smaller core, this effect would be reduced. Even on earth, the effect is not really noticeable at depths we can dig to. So the effect may not be present on the moon, ...


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I can't find any description of how the equation you cite is derived, so I can only speculate. With that caveat, I would guess the factor of 1.3 is the ratio of the rim diameter to the excavation diameter. The bolide will excavate an initial bowl shaped crater, and the diameter of this is the excavation diameter. Immediately after the impact various ...


2

This link explains it: The Earth experiences two high tides per day because of the difference in the Moon's gravitational field at the Earth's surface and at its center. You could say that there is a high tide on the side nearest the Moon because the Moon pulls the water away from the Earth, and a high tide on the opposite side because the Moon pulls the ...


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anyway, how likely is it the ice ages could be explained by the earth 'realigning' so that polar regions would migrate over the surface of the earth? How about zero? The geological evidence of the Ice Ages clearly says that, between the ice episodes, the ice did not move. It's just that the polar caps shrank. For instance, the extent of the last ice ...


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If you have a planet of mass $M$, then its self-gravitational binding energy is roughly $-GM^2/2R$ give or take a small numerical factor. So, for the Earth, this would be $-2\times 10^{32}$ J. Something colliding with the Earth, which has a similar mass and size, would do so at velocities of tens of km/s at least. I think the minimum closing velocity would ...


1

Anything over 500 miles in diameter, give or take is almost always sphere-shaped, the primary variation being rotation speed, which can give a flatness to the object, for example, Jupiter is visibly flattened by it's high rotational speed. The problem with building a strange shape by very large collision is that the heat generated in a collision of that ...


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Physical things (solid, liquid, gas, plasma) both absorb and emit energy in the form of electromagnetic radiation of a wide range of frequencies. How fast they radiate and the strongest frequencies of radiation depend on the absolute temperature. How fast they absorb depends on the temperatures of objects around them. Therefore, the net intensity (energy per ...


1

Whatever force you like. Since force just determines the rate change of velocity, you can use a massive force for a trillionth of a second or a tiny force for a long time period. However, if changed the velocity of the Earth (relative to the Sun) of 1 m/s, you'll would cause an impulse – change in momentum – on the Earth of $5.972×10^{24}$ kilogram-meters ...


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Force is just a product of mass and acceleration and to calculate the force required to move the earth is quite easy. Though incredibly difficult to produce. IF we take the mass of earth as 5.9736 x 10^24 kg. And the acceleration 1 meter per second per second. And that is where you are wrong when you said 1 meter per second which is the unit for speed or ...


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Your question makes sense if you replace "force" by "impulse". And that impulse is simply the mass of the Earth times $1 {\rm m\,s^{-1}}$. Moreover, you need to make the qualification: how much impulse is needed to change the Earth's motion state so that it is moving at $1 {\rm m\,s^{-1}}$ relative to its motion state now. According to the estimates of this ...



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