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We'll first write out the equation in its "full" form. \begin{equation} m\vec{a} =m\vec{g} -cv^{2}\hat{v} \end{equation} We will be using the convention that down is negative. We will also be assume we are dealing with two dimensions here. Therefore, we can expand this out as \begin{equation} ma_{x}\hat{x} +ma_{y}\hat{y} =-mg\hat{y} -c||\vec{v} ||^{2}\hat{v} ...

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Welcome to Physics SE Alexa! In a solid molecules are connected to each other by Wan der Waals forces. Those forces are attractive but when the molecules come too close to each other they become repulsive. All molecules will try to move down due to their weight ( or force exerted from earth on them). Molecules located near the ground will eventually ...

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The problem comes from the sign of the drag term. The first form of the equation you have is not exactly correct. When you say $$m \dot v = g - c v^2$$ there is an assumption that $v$ is positive, so the negative sign in front of $c v^2$ ensures that the drag for opposes the motion. Often, sign conventions can get confusing when replacing vectors with ...

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While other answers are not wrong, I'd like to add that the main reason why someone's feet can't stay on the ground in free fall is that any time there is a slight pressure or bump between feet and floor due to body movement, the contact force will push the feet upwards. Since no other forces are involved, the feet will keep floating away from the floor. ...

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Can a air tight tube be used to siphon water very high vertically as long as the end location is lower than the start? tl;dr: No, the water will spontaneously boil in the vacuum produced and since unlike nearly-incompressible liquid water, the vapor is a compressible fluid, it will no longer allow the siphon effect to continue. Atmospheric pressure is ...

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I have raised a similar question here and finally I ended up answering it. My answer is on the derivation of the $2^{nd}$ law which is the lengthiest of 3. The bonus is that I have done the complete proof using cartesian co-ordinates, so even a high school student with calculus knowledge can understand it.

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When we analyse the situation from a freely falling elevator, an non-inertial frame of reference, we need to apply inertial (pseudo or fictious) force in order to apply the familiar Newton's laws of motion. Here, the pseudo force acts in the upward direction and cancels the weight of the person. The net force on the person is zero. The following diagram ...

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Yes, they would stay in the same place... Except for the fact that from the elevator's frame any slight push upwards will propel you and you can't stop that... Like if you were in a space station. In the second example you gave you can clearly see that all the people push upwards which causes them to float.. if you stayed perfectly still you would still be ...

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I believe, the volume of ball immersed in water on the moon will be greater than that on earth. It's because, when you say it is partly immersed, buoyancy due to air must also be considered and there is no air on moon. So, the buoyant force experienced by the body on moon will be less, making it get a bit more submerged. (Sorry for my english though, it's ...

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1) If we can visualize angular momentum in 3D as a vector, how should we think about its counterpart in 4D, a tensor? And what about the law of conservation of angular momentum. In general, angular momentum is a geometric object called a rank $2$ differential form. To oversimplify a bit, you can think of it as a plane, along with a signed magnitude. In ...

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Does the object have acceleration in the tangential direction of the path? Yes, except for in the very beginning. The acceleration is straight downwards (gravity) so for any motion that is not horizontal there is a tangential acceleration component. there is acceleration only in the vertical direction, and the speed is constant in the horizontal ...

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Assuming air drag (and other forms of friction) is absent then the only force acting on the object is gravity. This force causes the only, vertically downward, acceleration acting on it, that is $\vec{g}$. Put slightly more mathematically, suppose the projectile is fired with an initial, horizontal velocity of $v_0$. The velocity vector $\vec{v}$ of the ...

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Take the Sun(mass M) and the planet(mass m) to be point masses. Let the Sun be fixed at the origin and the planet be moving in the x-y plane, initial velocity of the planet be $v_o\hat{j}$ and initial position of the planet be $r_o\hat{i}$. At any given instant, let the planet's position, velocity and acceleration be $\vec{r}, \vec{v}, \vec{a}$ respectively. ...

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When you calculate the orbital eccentricity in terms of energy and angular momentum, the same formula applies for circular orbits, elliptical orbits, parabolic trajectories, and hyperbolic trajectories. That formula is $$e=\sqrt{1+\frac{2\epsilon h^2}{\mu^2}}.\tag1$$ Here $\epsilon$ is the specific orbital energy (the energy -- kinetic plus potential -- ...

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I can't see where you get 20 million light years from. They are 2.5 million light years apart now, and will collide in about 5 billion years time. This suggest perhaps that when they formed around 12.5 billion years ago, they might have been, at most, 9 million light years apart. Both galaxies are of mass $\sim 2\times 10^{42}$ kg. If they were 9 million ...

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We don't actually know that Andromeda is on a collision path. It is coming closer, but in the absence of an accurate determination of proper motion, that does not imply collision. Gravity cannot (on its own) explain the observed randomness in peculiar velocities of galaxies. Galaxies were formed from gas clouds. The conclusion should be that peculiar ...

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they could be 20 million light years away at some point yet their gravity started to attract each other? Gravity has infinite range. It gets weaker and weaker as you go further away, but there isn’t a cutoff point where it goes to zero. It’s an “inverse square” law and $1/r^2$ is nonzero at arbitrarily large $r$.

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A large number of small marbles confined to a jar can be treated as a fluid when you are shaking it vigorously. When marbles of different densities are present, Archimedes principle applies here, and the denser (heavier) marbles tend to go down. Also, it is energetically favourable for the system to have its more massive components lower down. Unlike fluids,...

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Such envelope already exists around Earth, though it's full of holes. We call it clouds. Although it's not separated by vacuum but just by air, and that air plays a pivotal role in supporting it. A more dense envelope is not theoretically impossible, but there would have to be some effect that keeps the water from falling down. If the particles of water ...

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How can a resting object like earth have so much (infinite) energy to attract objects? Does this energy ever fades away, assuming no other external factors? As @Alure explained to you, gravity is a force not energy. Energy can be stored in and retrieved from the gravitational field. When you raise an object the energy you expend is stored as ...

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In physics, you only need energy if you have a force and a displacement. Just exerting a force alone does not require energy. This comes from the definition of work. Therefore gravity is not free energy; it's just a force. Your drone does not need to spend energy battling gravity; it needs to spend energy to propel air molecules downwards because that's how ...

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Actually you can. Both are one and the same. $F_{moon}=m_{moon}g$ $F_{moon}=\frac{GMm_{moon}}{r^2}$

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According to Abhinav's suggestion, $a=v\frac{dv}{dx}$. Rearranging the terms and integrating both sides gives us $$\int a dx=\int vdv=\frac{v^{2}}{2}$$ Inserting $a(x)=\frac{G}{(1-2x)^{2}}dx$ gives us $$v=\sqrt{G(\frac{2x}{1-2x})}$$ Inserting this into $dx=vdt$ gives us $$\frac{dx}{\sqrt{\frac{2x}{1-2x}}}=\sqrt{G}dt$$ Integrating the left side in Wolfram ...

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Two body problem: $R(\tau)=0$ with: $$\vec{r}=r\,\begin{bmatrix} \cos(\varphi) \\ \sin(\varphi) \\ \end{bmatrix}$$ Kinetic Energy $$T=\frac{1}{2}\mu\left(\dot{r}^2+r\,\dot{\varphi}^2\right)$$ where $\mu=\frac{m_1\,m_2}{m_1+m_2}$ Potential Energy $$U=-G\,\frac{m_1\,m_2}{r}$$ The equations of motion $$\mu\,{\frac {d^{2}}{d{\tau}^{2}}}r \left( \... 6 NOTE: This answer is for the OP’s original question, which was about two masses. The question was completely changed after this answer was posted. To get an ellipse with semi-major axis a, the initial relative speed should be$$v=\sqrt{G(m_A+m_B)\left(\frac2r-\frac1a\right)}.$$This is known as the vis-viva equation and is an expression of energy ... 0 You have an extra time derivative in your expression for Q, there should be just one. (Your present expression for Q vanishes, for the same reason you have described in the question). Having made that alteration, if we demand that \frac{dQ}{dt}=0, we will need to substitute the equation of motion. Upon substitution, this turns out to be precisely the ... 3 I am assuming you are talking about an object released from rest sliding down an incline. There are two forces acting on a body sliding down a frictionless ramp: gravity and the normal force supplied by the incline that is perpendicular to the incline. Both are important for the object to move down the incline. Gravity: Without gravity, the object would ... 4 Here is an algorithm you can implement. Let the small orbiting body have mass m. You know its initial position and velocity vectors, \mathbf{r}_0 and \mathbf{v}_0, in an arbitrary Cartesian coordinate system. I’ll assume that you want to treat the “significantly heavier” body with mass M as being stationary and consider it to be the origin of your ... 1 The rotation of planets comes from the fact that angular momentum is conserved. When the planets are formed the individual parts have some rotation, whose total effect is added up to give the planet some net rotational angular momentum, which is now conserved/it changes very slowly. Now there is a very low chance that the net rotational angular momentum ... 0 Once there is a force, there is also an acceleration (by Newton's second law). Once you have acceleration, you don't expect things to be stationary. After all, acceleration is the second derivative of the position x with time, d^2x/dt^2. 0 It depends. Usually the skydiver starts from zero vertical speed, the resultant force is downwards and his/her speed increases (ie there is acceleration) until terminal velocity is reached. At this time the drag force equals the skydiver's weight. However if the skydiver then opens the parachute the drag force suddenly increases so that the resultant ... 2 The Newton law of universal Gravitation is given by$$ F = G \frac{m M}{r^2} $$For two bodies of mass M and m. When describing gravitational effects near the surface of the earth, its common to use the approximate formula F = m g, where g = G M/R^2, with R being the earth´s radius and M the earth mass. -3 Two thoughts: Reduce the mass of the planet Slow the rotation of the planet I like these way better than all the trains, planes, and magnet answers. The reason other planets have lower gravity is because they are lighter. Just like moons. To note though: Rotation of a planet does not affect the gravity, but the rotation of a planet reduces the ... 1 Orbital mechanics are weird. So, (hypothetically) we have the Earth travelling in a perfect circular orbit around the Sun. The radius of that orbit, the velocity of the earth, and the mass of the Sun are all fixed, and satisfy the relationship cited in the question. Now the Sun loses a small amount of mass. The Earth is still traveling at the same ... 1 Welcome to Physics.StackExchange! Let M(t) be the solar mass and m the Earth mass. Assuming circular orbit with radius r(t), The gravitational force between them is$$ F = \frac{\gamma mM(t)}{r^2} = ma = \frac{mv^2}{r} \Rightarrow \sqrt{\frac{\gamma M(t)}{r}} = \frac{dr}{dt}, $$where the orbital velocity is  v = dr/dt. Further assume the mass of ... 1 Consider Kepler's Third Law:$$T^2=\Big(\frac{4 \pi^2}{GM}\Big)r^3$$where T is the orbital period and r the orbit radius (for a circular orbit). So:$$T^2 \propto r^3$$where \frac{4 \pi^2}{GM} is the proportionality constant. Since as it depends on M, the ratio \frac{T^2}{r^3} will vary accordingly. 2 The only feasible way I could think of is Reduced-gravity aircrafts. These are used for training astronauts and conducting experiments in reduced gravity environments. The following image shows the trajectory of such an aircraft: Image source: Purdue University The path looks like a downward opening parabola. During the ascending phase, contents in the ... 2 The cheapest solution would be to use magnetic levitation principle as it is used in some high-speed train systems. One needs to cover a ground with a permanent magnetic field coating and to use some kind of boots with repelling magnets for levitating from the ground while walking. The only problem - stabilization. Levitating train has no such problem cause ... 10 Edit I am adding new headings for each section of possible solutions. Centrifugal Force Suppose that the planet is not revolving at all then the effect of gravity can be reduced by moving around the planet in a high speed train (assuming that by reducing the effect of gravity you mean the measured weight). How? When you move with speed v around the ... 0 If you mean by your question the measured acceleration of a free falling object then the lowest acceleration is almost certainly on some high peak near equator because 2 factors are in favour of that: 1) Earth is ellipsoid so places on equator are more distant from Earth center then places on poles (and gravity decreases with distance). 2) Centrifugal (... 5 Here is a map of the gravity field of earth given by NASA Earth's gravity measured by NASA GRACE mission, showing deviations from the theoretical gravity of an idealized smooth Earth, the so-called Earth ellipsoid. Red shows the areas where gravity is stronger than the smooth, standard value, and blue reveals areas where gravity is weaker. So one would ... 2 Proceed as follows. First accept that the notion of a point charge (with a finite amount of charge but no volume) is itself a mathematical abstraction which may or may not be useful or appropriate in dealing with any given piece of analysis. Next, treat a small charged region, say a sphere, with finite charge density and finite fields. Finally, let the ... 7 The divergence theorem is stated (using vector calculus notation)$$\iiint_{V} \vec{\nabla} \cdot \vec{F}\ dV = \iint_{\partial V} \vec{F} \cdot \vec{n}\ dS$$For some C^1 vector field \vec{F}. If we consider, say, the electric field of a point particle :$$\vec{E} = \alpha \frac{\vec{e_r}}{r^2}$$for some constant \alpha, then E isn't ... 1 The body itself does not need to directly hit the ground. Its the impact force transmitted to the body that counts. The table legs would have to absorb enough of the impact energy so that the impact force transmitted to the body does not result in a fatal injury. This is the reason that automobiles are designed with crumple zones to absorb energy in a crash ... 1 I don't know the stress a bone is able to withstand before fracture. However, a simple approximation yields that it is possible to survive this fall: Let's assume that we are falling h_1=29m and that we deaccelerate constantly during the last h_2=1m. In this scenario we obtain the energy E = m g h_1 during the free fall and experience the ... 1 The reason for this is that the formula$$F_G = G\frac{m_1 m_2}{r^2} technically only gives the gravitational force between two objects that are of point size, i.e. that have size zero. When $r = 0$, that is saying that you have two infinitely concentrated masses infinitely close together. Does this sound like a good description of the situation of an ...

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We have the power to overcome this near infinite force of gravity because the electromagnetic forces generated by our muscles is much much stronger. Every cell in our body burns on the order of 1 to 10 million molecules of ATP every second. That’s when those cells are at rest. During periods of high intensity, that ATP burning increases 1000 fold in ...

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