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The vis-viva equation is commonly written like this: $v^2 = GM(\frac{2}{r} - \frac{1}{a})$. For $r=a(1-e)$, (and a solar mass star) that becomes $v = \sqrt(GM_{solar}(\frac{2}{a(1-e)} - \frac{1}{a})) = \sqrt(GM_{solar}\frac{1}{a}(\frac{2}{1-e}-1)) = \sqrt(GM_{solar}\frac{1}{a}(\frac{1+e}{1-e}))$. The derivation of this formula is not at all trivial and can ...

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I find this comparison of tangential velocities on Wikipedia very confusing. … According to it, the tangential speed of Earth's surface (465.1 m/s) is different from the tangential speed required to "orbit" at Earth's surface (7.9 km/s). That might be, but they have explicitly elaborated "… Earth's own rotation at surface (for comparison— not an orbit) …...

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An interesting corollary to this question: if the ground is not in orbit, how does it move (roughly) in a circle? If we model a section of ground as an isolated particle, it's clear that in order to move in a circle despite having a relatively low tangential velocity, it would need to have an ongoing force being applied to counteract the direction the ...

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1. Is material on Earth's surface not in free fall around Earth's center? No. Material on the Earth's surface -- or inside it -- is not in orbit, and so is not in free fall. You can temporarily put yourself into an orbit (and thus into free fall) by jumping up into the air, or jumping off a higher surface. When you do this, you are briefly in a very ...

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Imagine you are in orbit around the earth, several 100 km upwards. What happens when you slow down? That's right, you fall down until some force stops your fall. That force is the pushback from the ground. So next imagine: What happens when you throw a ball in the air? It falls back down to the ground. So it follows, that the ball is too slow to be in orbit....

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Judging by the fact that after cubing the distance you have ended up with an exponent of 1021 (21 = 6×3+3) at the first point you have substituted the numbers into the equation, you appear to be assuming that 1 km3 is 1000 m3. However, if you write the same linear quantity in metres and kilometres and cube them, you can see this is not ...

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Here's how I would do it. I'd convert everything to a single, standard set of units as recommended in the comments, and also stick to one digit before the decimal in scientific notation: $$T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}$$ Using all numbers in the same units: $r \ 4.5 \times 10^{11} \ (m)$ $G = 6.674 \times 10^{-11} \ (m^3 \ kg^{-1} s^{-2})$ $M ... 3 After what I recall from astronomy class ... for the gas cloud to form a star it has to overcome some "difficulties" - forces that restrain further shrinking. In the first place the molecules need to move very slowly .. only cold clouds can become stars - otherwise they couldn't form regions with higher mass concentration that would start the entire ... 8 Two massive bodies orbiting each other form stable orbits. This is called the "two-body problem." Add a 3rd body to the system and the results are unstable orbits. It's akin to the motion of a singe pendulum. A single pendulum swinging back and forth is a linear system with motion that is repeating and predictable. Similarly two bodies orbiting each ... 30 Collapsing gas clouds fragment into multiple cores because the Jeans mass, that determines the minimum mass that becomes gravitationally unstable to collapse, becomes smaller if the cloud is able to contract without heating up too much. i.e. $$M_J \propto T^{3/2} \rho^{-1/2},$$ where$\rho$is the cloud density. Thus if the cloud density can increase but ... 1 Partial answer for now. tl;dr: It will be an "exercise in futility". This is a great question! With tens of thousands of very similar satellites in LEO the number of occultations that will occur from a given observatory will be much larger and better characterizable than with the current diverse population, so "if it were useful it would have been tried ... 2 That was likely the latest StarLink satellite constellation. Note that usually it is important to give an exact time and place to answer questions like this, but in this case the answer looks fairly obvious. It really does look odd to see the constellation fly overhead (later it will disperse into separate orbits and not be very conspicuous). 1 I think that at the moment, all you have available for the pair's orbit around their star is a snapshot of their two distances; 964 and 989 seconds, or about 289 and 297 million kilometers. Your snapshot doesn't have any velocity information, so I think that there's no way to calculate the mass of the star from that data. Since the first planet is 2000 times ... 4 In the comments of the other answer, the question came up whether decay of the orbit would limit the time amount of time dilation. The answer is of course yes. But by how much? This question can be answered using some modern results for the modelling extreme mass -ratio binaries in the limit of extremal spins. The answer will depend crucially on ratio of ... 6 Yes, but not very likely. The closest orbit that does not require constant expenditure of energy to maintain it is the prograde equatorial ISCO. For a Kerr black hole the time dilation factor on this orbit is $$\frac{dt}{d\tau}\approx \frac{2^{4/3}}{\sqrt{3}(1-a/M)^{1/3}},$$ which at the astrophysically likely Thorne limit$a = 0.998M\$ gives a dilation of ...

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