Hot answers tagged

8

Quick summary of what a "Caplan" thruster is: A Dyson swarm collecting sunlight, shooting it back at the Sun to stir up mass as solar winds. Some system to collect that solar wind. Fusion reactors, using the helium from the collected solar wind. A fusion product jet of oxygen-14 pointed into space. A hydrogen jet pointed back at the Sun. Caplan's ...


6

Reaching your "falling" trajectory means leaving the Earth with a $v_{\infty}$ of 29.8km/s. From low Earth orbit, that's a delta-v cost of 24.0 km/s, which is quite steep. If you waive the requirement of falling straight towards the centre of the Sun, and consider hitting the Sun good enough, the delta-v cost goes down to 21.3 km/s. Your worry ...


5

How do solar flares and coronal mass ejections create proton storms? The proper answer to this questions fills multiple books and they all still end with lots of questions, but I will try to give a summary of what they say. First, solar flares are just the localized enhancement of electromagnetic (EM) radiation (usually dominated in UV and x-ray bands) ...


5

Let's try to make some quick and dirty estimation for the upper limit in the case of Earth. (And in the end figure out that we have an actual measurement of this...) First, the bending power of the atmospheric lens. It's not uniform as @uhoh mentioned in his answer. But, for rays passing just above surface we can get a accurate number. During sunset ...


5

The Sun is a large ball of plasma and does not rotate at a uniform speed. It rotates fastest at the equator, where it takes 24.47 days to make one revolution and up to 38 days at the poles. It takes an average of 28 days to do one rotation [1]. You can calculate the rotation rate of the Sun vs latitude with the following equation [2]: $$\omega=A+B \sin ^{2}(\...


5

Yes. It was shown and described in the NASA press conference at about minute 31: And here is a close up of the Gnomon from the latest press conference showing much more detail:


4

Yes, according to mastcamz.asu.edu . The text there calls it a 'calibration target' but makes it clear that it's a Marsdial chock-full of inscriptions as with Curiosity's Marsdial.


3

Doesn't this mean that the Moon's orbit intersects the Earth's orbit around the Sun at the orbital nodes? Not in the general case. Consider an arrangement where the lunar nodes are oriented towards and away from the sun. The moon will be above (or below) the ecliptic when it crosses Earth's orbit ahead of the Earth, and below (or above) the ecliptic when it ...


3

I don't know how exactly to quantify ...most spectacular, yet realistic, trajectory... but ...it has to be aiming the very center of the Sun as much as practically possible... is easy. You ask for companies to bid on a $C_3 = v_{Eorb}^2$ launch. $C_3$ is the reduced kinetic energy (per unit mass) of an object leaving Earth in a geocentric frame. It's how ...


3

There is a detailed review of space weather effects on humans in space by Townsend [2021]. They highlight several solar energetic particle (SEP) events that would have exceeded 30 day short-term organ damage limits from recent observations, all of which are likely to be weaker than the Carrington event. They also provide the dose limits for blood forming ...


3

It is absolutely possible to be still with respect to the center of the sun. This can be thought of as hovering over the sun at perhaps a very large distance. Thrust is required to avoid falling towards the sun and also being perturbed by other objects. It is bad to call this "standing still in space", since the center of the sun is an arbitrary ...


2

Satellites like DSCOVR are used to predict solar storms. No they are not. Spacecraft cannot predict solar storms. They can inform a user that something is coming toward Earth, but they cannot predict the onset of solar phenomena. That is, spacecraft are capable of providing a warning but they cannot predict anything (I will distinguish that users ...


1

Partial answer, because this is a really hard question! We can learn how lenses work using the thin lens approximation where the amount that a ray is bent (the angle, or the tangent of the angle) is proportional to the distance from the center of the lens. For example if my lens bends rays 1 cm away from the axis by 0.57 degrees (0.01 radians) then it's ...


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