New answers tagged

1

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 ...


4

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 ...


0

You asked "how do I sum more than two vectors". In general, you can write your vectors as $(x_i, y_i)$ where $i$ counts from 1 to 4 in your case. Then the sum vector is $$x_{sum} = \sum x_i\\ y_{sum} = \sum y_i$$ The forces you have are all pointing towards the center, which means that for each of the forces, you will have either $x_i=y_i$ or $x_i = - ...


0

well you see the lift has got to do with the mass. Because of the design of the wings of the airplane the speed of wind going above the wing is more than the speed of air going from the bottom of the wing. Therefore from the Bernoulli's equation the pressure above the wing is less than the pressure below the wing. this pressure difference causes the lift. ...


1

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

Note that pendulum motion is only sinusoidal for small angular displacements; as you increase the amount of swing the harmonic approximation fails. Lagrangian mechanics gives you a handle on all of the cases.


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 ...


0

Your weighing machine measures gravitational force which is your real weight(W=mg) on earth at that point where you are measuring. And your mass would be according to formula W=mg.


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F = MA. The force required to lift a plane is mass times acceleration. Acceleration on earth is around 9.8m/s/s. The mass increase will therefore increase the amount of force it's applying, and therefore the force required to cancel it out. Two balls of the same shape but one has more mass, it will fall at the same speed as the other, but will press harder ...


1

There is always air resistance, unless you are in space. If there was no air resistance, a bullet would land at the same speed it was shot at. On Earth, a spent bullet is rarely lethal. In big wars (like World War II) it was common for soldiers to get hit by spent bullets. Normally they will hit you and fall away, causing just a bruise, An unlucky hit ...


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 ...


0

You can't add forces on different objects to get an acceleration of one object. All the forces should act on the center of the sphere to have effect on breaking the sphere. All the three forces: the tidal force $F_{T}$, the gravitational force $F_{g}$ and the Tension force because of the rope pulling the sphere $T$ act at the center of the sphere for the ...


5

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 ...


0

It is not so clear what you mean by "infinite", but maybe you mean to ask why gravity has infinite range. It comes from the fact that the force of gravity decreases as $1/r^2$, thus slow enough for it to be felt by objects at the other end of the Universe. The same property is seen in the electrostatic Coulomb's force that decreases as $1/r^2$ as well. This ...


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 ...


0

$\newcommand{norm}[1]{\lVert #1 \rVert} \renewcommand{vec}[1]{\pmb{#1}}$Let's take a mathematical point of view. Let the mass of the objects be $m_i$ for $i=1,2$ and the length of the rod $l$. The total torque with respect to the center of mass (COM) is given by: $$\vec \tau \cdot \vec e_z = \vec e_z \cdot \left[ \vec r_1 \times \vec F_{g1} + \vec r_2 ...


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 ...


0

The acceleration of the points $B$ and $C$ will be same assuming that the Earth is not breaking apart, i.e., the acceleration of all points on the Earth will be same. If you think of the Earth as a very strong rod, only the Tension in the rod varies so that the entire rod accelerates with the same acceleration as that of the center of mass of the rod. So, ...


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 ...


1

C. Newton's Shell theorem states that if you're inside a hollow spherical shell that has matter even distributed in the shell, then the net gravitational pull you'd feel would be 0. This is because the attraction to the small near part of the shell is equal to the attraction to the large far part of the shell.


0

In physics, you always have to know the meaning behind the formula. The formula itself is meaningless, it's just a way of writing the meaning in a short and universal form. Even if you have two variables that describe the same quantity, but in a different context, equating them will probably lead to nonsense. In this case, E is precisely the REST energy of ...


0

I think it would help for you to first understand what does this E in E=mc^2 means. Yes it means energy. But more specifically, E in this equation is the energy you can generate by converting m amount of rest mass into energy. So the idea this equation describe is that you can convert rest mass into energy, or you can convert energy into rest mass. It is ...


0

The mistake is that bodies at rest don't have zero energy. The relativistic energy is given by $$ E^2=p^2c^2+m^2c^4 $$ where $p$ is the body's momentum and $m$ its rest mass. Hence, if the object is at rest, $p=0$, so $$ E=mc^2 \rightarrow m=\frac{E}{c^2}\neq 0 $$ and $$ W=mg=\frac{E}{c^2}g\neq 0 $$


0

There's a new equation which serves as a correction to the previous one, ${ E }^{ 2 }={ { m }_{ 0 }^{ 2 } }{ c }^{ 4 }+{ p }^{ 2 }{ c }^{ 2 }$ Or as previously stated in a comment, objects at rest do not have $E=0$ or $m=0$. Misunderstandings including this have made people reluctant to talk about relativistic mass. If the given system is at rest, ...


1

Well you can find approximate values by reasoning but for the exact value you'll need at least to know limits. Let's reason for a moment as you are proposing, but slightly different. Since the plane is infinite you can consider the point in the center of the plain, and at distance R from it. Now think of the plain as a collection of rings of very fine ...


0

Between points A and C The earth can be considered as a wall and thus to break off the 'wall' a relative acceleration with respect to the wall is a must , If the tensile strength is 10N and mass 10 kg then the rope must move at a relative acceleration of 1m/s^2 or Acceleration(a) - Acceleration(c) = 1


0

I have been using Orbital Mechanics for Engineering Students by Howard Curtis available on Google Books has a great derivation of the Universal Variable formulation, see chapter 3.7, as well as step by step algorithms and example Matlab code for calculation. I have been using it as a basis for a python program I am developing (as a hobby) and highly ...


1

The number of tides stays just about the same, as the earth turns under the tidal bulges. On a water earth we would have two tides per day. Local landforms impact that greatly, varying from place to place. The tidal force is the difference of the moon's gravity on the near and far sides of the earth. It falls off as $\frac 1{r^3}$, so the if the moon ...


0

The tides will get smaller as the moon moves away having less gravitational effect. As long as there are tides there will be two per day because this is based on the Earths rotation not the moons distance. As the Earth spins the tides rise on the side facing the moon and the opposite side because of gravity. Its these tidal forces along with the Earths spin ...


1

Just imagine there would be no moon at all. Obviously there would be no tides due to the moon. As Dr. Chuck pointed out in the comments, there would still be tides due to the sun, but the tidal forces of the Sun are only about 46% of the moons tidal forces (taken from german Wikipedia about Tides: https://de.wikipedia.org/wiki/Gezeiten). So the farer the ...


0

Looks like you are thinking that motion should be along the line on which the Net Force lies as well as in the direction of Net Force. This is not necessary. Force is only supposed to change the velocity of the body. The resulting velocity is not necessarily/always along the line of acceleration i.e, change of velocity. The Centripetal force create ...


1

In order to arrive to this answer you have to assume that the size of the planet and star is negligible compared to the size of the orbit of the planet. When solving this question you have to understand what each symbol stands for: $T$ stands for the orbital period of the planets orbit; $r$ stands for the semi-major axis of its orbit; $GM$ can be combined ...


1

I think what is missing is the definitions of $F$ and $V$. Consider first the definition of potential energy. The potential energy at a point relative to another point is the work done by a external force (eg the force exerted by you on the mass, $\vec F_{my}$) in taking the mass from the first point to the second point. That force which you exert on the ...


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.


1

I mean, if you believe what you've just written, $$\theta(t_2)-\theta(t_1)=\theta(t'_2)-\theta(t'_1)$$ Then I think you have already proven it, with a little more formal calculus. First rewrite that expression above as $\Delta \theta(t)=\Delta \theta(t')$, with the prime indicating the other interval. If $$t_2-t_1=t'_2-t'_1,$$ Then just set $\Delta ...


1

The mentos thrown faster than terminal velocity will experience greater drag until it once again has terminal velocity (note - terminal velocity is only ever approached asymptotically, but we can decide to call "close enough"="equal") If one thing is going faster than another for a while, after which they are traveling at the same speed, then the faster ...


1

The terminal velocity is by definition the velocity for which the gravitational force is equal to the friction force of a falling body. As the friction force grow with speed (linearly for laminar flow, quadratically for turbulent flow) an object going faster that the terminal velocity will have his friction force bigger than the gravitational force and thus ...


1

There's a lot of text here, hopefully it's not too much and makes at least a little sense. $\ddot\smile$ There are several units for gravity, depending on how you model gravity and what you're looking to calculate. Energy, in Joules ($J$); or spacetime distortion, but I don't know the units, both measure the "absolute" amount of gravity near a point, and ...


-1

using G ,u can find magnitude of acceleration of any Celestial body.Earth has different radii of curvature at poles and equators. so magnitude of gravity is different at different earth positions


1

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


-1

Gravity (g), is the acceleration that the Earth imparts to objects on or near its surface. It is measured in metres per second squared (m/s2). The gravitational force on Earth is different from that of other planets as the force is determined by the size of the planet, thus why when astronauts walked on the moon they were almost floating. Gravity also ...


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 ...


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 ...


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 ...


1

Your final question very much correlates with a famous thought experiment.If the Sun was suddenly removed the planet s will still continue to stay in orbit. For 8 minutes and 20 seconds. This is because the speed of the space time fabric or simply putting gravity travels at the speed of light. That is, the earth will be devoid of sunlight and will move ...


-2

A coarse correction would have to be made because of the difference in the curvature of space in that section of space when a planet is deleted. The sun holds the planets in orbit and they would fly away as if the string to a tether ball was cut to all the planets.


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

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 ...


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 ...


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 ...



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