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Calculating the sum of the interior angles precisely woud be a big task as we'd need to compute the trajectory of the light ray and there isn't a convenient analytic expression for this. However we can easily calculate an upper limit for the interior angles. The key fact we need to know is that the deflection angle $\theta$ of a light ray in the ...

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Now, one g is equal to the acceleration due to gravity by Earth on its surface. Force by washing machine is the centrifugal force as clothes try to go along straight path but the resultant provided by the walls of container of machine which provides the required centripetal force. Now, according to Newton's third law of motion , clothes will also give equal ...

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The simplest formula for the centrifugal acceleration is $$a = r\omega^2$$ Here, $r$ is the radius which is 0.25 meters in your case. $\omega$ is the angular velocity which is $2\pi$ times the frequency $f$. Your $f$ is 1500 revolutions per minute which is $1500/60=25$ revolutions per second. In the SI units, we have $$a = 0.25\times 4\pi^2 \times 25^2 = ... 4 Actually you can go to the orbit of Jupiter with a \approx 2500 tonnes rocket and a 3 tonnes payload. From there you can use an ionic engine. A rocket launched from the equator of Jupiter that turns at 12.6~\text{km}/\text{s} needs just an increase in speed v = 29.5~\text{km}/{\text{s}}.$$v_{rj}:= 12.6~{\text{km}}/{\text{s}} \;\;\; R_j := ...

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Let's assume you mean that Earth now has the mass of Jupiter (as opposed to actually launching from the literal planet Jupiter - whole different question...). Then: radius of Earth = $6.4 \times 10^6~\text{m}$ mass of Jupiter = $1.9 \times 10^{27}~\text{kg}$ Escape velocity, $v_\text{escape} = \sqrt{\frac{2GM}{r}}$ This gives a value for ...

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Hey! The question keeps getting edited! Make up your mind! You asked about Mars originally, then edited the question. Actual, real Jupiter is flat out impossible. Does it have a surface to launch from? Who knows? What's the pressure at that depth? Can our probes even survive at that depth? Probably not? What if Earth had the mass of Jupiter? More ...

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Most of the energy of running is used to move the legs of the runner. This isn't very efficient because the legs are heavy and are being quickly accelerated and deaccelerated from the running speed. The leg muscles are very powerful but runners hit a maximum speed when most of the power of their muscles is used up moving their legs back and forth. Bicycling ...

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If you ignore air resistance, and assume the surface is hard and absorbs no energy that you need to care about, then the remaining energy must be absorbed as heat in your body (and shoes). If your temperature is steady then you will lose energy through heat radiation and convection at the rate you use it. There will be many different chemical and mechanical ...

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The Earth has an average density of about 5500 kg/m^3. For a small planet, the density would be pretty similar throughout. Therefore the force of gravity you would experience on a planet with Earth density is: Your Mass * 5500 * 4/3pi * r^3 / r^2 * 6.6723 * 10^-11 This is equal to about 1.456 * 10^-6 * Your Mass * Radius of Small Planet. (I know, if you ...

3

The problem is in your assumption that "we fall into open space" unless the planet is large enough. Even if there were no planet at all, we would not fall. In open space you just stay where you are - unless you are affected by some star or planet. In other words you will always drift slowly towards something or other. Now the earth's gravity is so large ...

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Lets just start with the cosmic velocity assuming you dont want to escape fully. $v \le \sqrt{\frac{2GM}{r}}$, where M is the mass of the planet and r is the distance to the center of Mass. We assume a spherical planet. We know the average density $\rho$ of our planet earth and since you will probably want to live on the new planet we will assume it has ...

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Let's do some rough estimation. Considering: CuriousOne's very good and fitting comment with regards to geometrical estimation given in question and the last but not least the human physiology ($10^{-12} \ W$ can hear only a full healthy man, with perception much less than a whisper and only on a frequency range improbable due to the diffraction) we ...

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To get a spacecraft to the Moon we normally use a Hohmann transfer orbit. The fuel is used in two steps: increase the velocity of the scapecraft to put it into an elliptical orbit with its apogee at the Moon. when the spacecraft reaches the Moon increase its velocity again to match the velocity of the Moon. The amount of fuel required is described by the ...

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Ignoring the difficulty with blowing a sizable piece of the moon off, there's 2 primary factors to consider. Distance and angular momentum. Distance The moon actually rotates. All tidally locked objects rotate, they just rotate, they just rotate in sync with their orbit. The Moon rotates once every 28 days, the same as it's orbital period. If the ...

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The answer to this question is fairly simple, yet complicated. The reason why planets are planet sized objects form a spherical shape is due to the immense force of gravity. Looking at the force of gravity: $F_g = G{(\rho v)^2 \over r^2}$ As something obtains more mass it generates a stronger gravitational field. If you decide to build a spacecraft as ...

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Stars are gaz and plasma, planet can be considered as fluid or deformable material (and have pretty liquid early stage), so spherical shape is natural. The question is, is it physically possible to have materials that keeps rigid (or could make rigidifying structures) up to these orders of mass. I don't know if the answer can even be known, as I'm not sure ...

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