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16

First remember that $k = \dfrac {1}{4 \pi \epsilon_r\epsilon_o}$ where $\epsilon_r \ge 1$ It is because a medium can be polarised by an external E-field. The dipoles so set up produce the external E-field produce an E-field in the opposite direction so the net E-field (the sum of the external and dipole produced E-fields) is smaller. Thus the force a given ...


14

The formula tells us that the force of gravity is equal to the mass of Object 1 multiplied by the mass of Object 2 multiplied by the Gravitational Constant and then divided by the distance between the objects, squared. It is implied, I think, that both objects have mass. Correct so far as it goes. Physics is the scientific discipline which gathers ...


9

In electromagnetism there are both positive and negative charges. Hence the force due to electric charges can be attractive or repulsive. Gravity, when treated as a classical force field, can only be attractive, there are not two types of "gravitational charge". What this means is that in electromagnetism, a given medium, may contain both positive and ...


6

Here is the proof: Please refer to the wikipedia page on eccentric anomaly for a diagram and a couple of intermediate formulae. For an ellipse with the usual formula $x^2/a^2 + y^2/b^2=1$, it is the case that $\sin E = y/b$, and also by studying the figure on the wiki page you can see that $\sin (\pi-\nu) = \sin \nu = y/r$. Thus the two results you wish to ...


5

The term "mass" is an intrinsic property anf any body, and doesn't depend on external factors. The term "weight" is a force, i.e. it measures how much a mass is accelerated. Your mass is $m = 65\,\mathrm{kg}$. Your weight on Earth, which accelerates you at $g = 9.8\,\mathrm{m}\,\mathrm{s}^{-2}$, is $$ w \equiv mg = 65\,\mathrm{kg}\times ...


5

Because it's an excellent approximation. The gravitational potential from a spherical mass $M$ of radius $R$ is, to second order \begin{align} U(r) & = -\frac{GM}{r} \\ & = U(R)+\frac{dU}{dr}(r-R)+\frac12\frac{d^2U}{dr^2}(r-R)^2 +O\left(\left(\frac{r-R}{R}\right)^3\right) \\ & = -\frac{GM}{R}+\frac{GM}{R^2}(r-R)-\frac{GM}{R^3}(r-R)^2 ...


4

The scenario you suggest is of course hypothetical, but in all cases you must conserve angular momentum and mass/energy. So for example: If you have a way of removing mass from a star in such a way that the mass disappears outside the orbit of the planets (in astrophysics this is accomplished simply by mass loss - either the star has a wind that expels mass ...


3

If at any time the speed of the planet in the reference frame of the star exceeds the escape velocity $\sqrt{2GM_\star/r}$, where $M_\star$ is the mass of the star and $r$ is the distance from the star to the planet, it will escape in a hyperbolic trajectory (or straight line if $M_\star\rightarrow0$). As noted in the other answers, the result of the ...


3

Gravity is bound by the speed of light, it's not instant. That has a very real implication: if you have two objects a lightyear apart, it takes gravity a year to notice and react. There's an implicit hidden mechanism involved in the question here: if gravity needs a trigger to start, then what starts the trigger force? It too would be bound by the speed of ...


3

This is known as the two-body problem of modeling the interactions of two bodies. More specifically, it is called the Kepler problem, as the objects interact via an inverse-square force - gravity. If we define some parameter $u$ as $$u\equiv\frac{1}{r}\tag{1}$$ where $r$ is the radius of the orbit at some angle $\theta$, then, using the Euler-Lagrange ...


3

The multipole expansion in its most general form reads $$ V(\mathbf r)=\sum_{l=0}^\infty \sum_{m=-l}^lQ_{lm}\frac{Y_{lm}(\theta,\phi)}{r^{l+1}}, $$ where the $Q_{lm}$ are the multipole moments of the system and the $Y_{lm}$ are spherical harmonics, and in this form it is applicable to bodies of any shape, and located at any point in space, so long as the ...


3

Unlike the centre of masss the centre of gravity is not an intrinsic property of a mass distribution. The centre of gravity only becomes distinct from the centre of mass when there is an external gravitational field that is non-uniform i.e. varies significantly across the distribution of masses. If the external gravitational field is higher on one side of ...


3

I think your problem is in how you understand the gradient. Define your z-axis. Lets say it points up. Now $$V(z) = mgz$$. Thus gradient of V(z), that is $$\nabla V(z) = mg \hat z$$. Here, $$\hat z$$ is an unit vector pointing up. So the force will be a vector of length mg, pointing down, because of the minus sign.


3

First, when calculating force, the unit produced is newtons (N), not kg. You can use kg(force) if you're willing to risk getting confused, and your calculation of the falcon weight shows that you did get confused. In a 1g environment, 1 kg of mass produces 1 kg of force, so there is no multiplication by 9.8. The gravitational attraction between any two ...


2

Do we feel less weight on surface of Mount Everest? (Or have I mixed some wrong values?) The answer to both of these questions is "yes". One would weight a tiny bit less on Mount Everest, but not as much less as the question poses. You have used some incorrect values and assumptions. If you use the numbers you yourself used to compute the gravitational ...


2

your model is the one object model , with 1 and only 1 fundamental particle. If it is a composite object, gravity plays between its components. Else, it is impossible to state , a thought experiment without observer having no meaning : Any observer will act as a second object and interact with the first. Any memorization process will need a physical ...


2

The gravitational force you would feel inside the hollow planet is zero. Let's prove it. Let's call the interior of the planet $P$ (it is an open ball in $\mathbb{R}^3$. 1) Gravitational force is conservative, which means that $F = - \nabla \Phi$, where $F$ is the gravitational force, and $\Phi$ is a smooth function (the potential). 2) By Gauss' Theorem ...


2

Yes the bullets can fall down and injure or kill you. In fact in countries were celebratory gun firing is possible people are often injured by falling projectiles. Shooting straight up is less dangerous than at an angle because the terminal velocity is much lower than the muzzle velocity of the projectile. When shooting at an angle some of the horizontal ...


2

I think you are puzzled because you are not distinguishing the gravitational potential produced by the system and the one acting on the system. The multipole expansion you link to, is performed on the expression for the potential produced by the system on a distant point. In this case the system has a mass center, but no gravity center can be defined for it ...


2

This experiment is bound to fail, because a particle underneath the large mass will still be feeling the force of mg of the earth. The attraction of the large mass will have an upward force of (m_1xm_2/r^2)xG . From the value of the gravitational constant one sees that the equivalent "g" for the attraction of the levitating mass is very small for 340 tons ...


2

Assuming that the twin star system is isolated, there is no external force acting on it so the center of mass should not move if the system was at rest initially. Centre of mass doesnot move Centre of mass moves although no external force is applied and the system was at rest initially


2

In the example of the incline plane that you have provided, the mass does not affect the speed, because the only friction force present is proportional to the object's weight. However, oftentimes significant dissipative forces are proportional to the velocity of the object --- for example, if an object if freely falling through a viscous fluid. You could ...


2

The motion of a comet can be understood in terms of a couple of principles. First - Inertia. Newton's first law states that an object will remain at rest or in uniform straight line motion unless acted upon by an external force In other words - assuming that "something" had given a comet a velocity, it will keep going unless something changes that. ...


2

The problem is air. If something is lightweight compared to its surface area, like a feather, snowflake, dust mote, animal hair, it cannot fall as fast as a brick because air resistance slows it down. Even a brick or a bowling ball eventually reaches a maximum speed when falling in air. Any discussion of this has to assume (ideally) no air resistance, or at ...


2

It is true that the heavier plane needs a greater lift and this is seen in practice. If two planes at equal altitude loose power at the same time and one weighs more than the other they will be able to glide the ...... same distance! One of them descends faster than the other but it glides forward faster to generates more lift. It seems odd, but one ...


1

The basketball and the medicine ball are both round, and though medicine balls can be a little larger than basketballs, the surface area and shape that each presents to air resistance is roughly equivalent. The major difference is weight. Basketballs weigh in the neighborhood of 20 ounces, but medicine balls can weigh up to 24 pounds. The experiment you ...


1

As many here have noted, Earth poses a problem for your experiment. So, let's remove it$^\dagger$. You now have a freely floating rock in space of mass $M = 3.4\times10^8\,\mathrm{g}$ and radius $R = 330\,\mathrm{cm}$. The escape velocity of the rock is $$v_\mathrm{esc} = \sqrt{\frac{2GM}{r}},$$ where $r$ is the distance to the center of the rock, i.e. ...


1

None of this is done correctly. There is no integration required at all. The "equilibrium" force balance on the falling object is simply $cv^2=mg(r)$. So, if g is treated as a function of r, the "terminal velocity" varies with r, and the local value of r should be used in the equation $v(r)=\sqrt{\frac{mg(r)}{c}}$. But, how can we call it a terminal ...


1

Here's the way I understand it. Someone will correct me if I'm wrong I'm sure. Using the earth as an example, and assuming a constant mass density, for an object outside of the earth’s surface, it appears as if all the mass of the earth is concentrated at its center. This is a form of Gauss’s law for gravitational objects. If you could go to the exact center ...


1

You need to use energy because the value of $g$ varies with height. So something like $\frac 12 m v^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h}$



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