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1

That value for the mass of Mercury is not correct. Check with Wikipedia. The correct value is $3.3\times 10^{23}$ kg.


-2

Let's consider what can in principle could be calculated about G, given the fact that G is a dimensionfull constant whose value can be chosen completely arbitrarily, as made clear in the other answers. So, let's reformulate the problem. For an intelligent lifeform living on a planet, we can consider a unit system such that G is of the order of unity when ...


5

No, as commented on above. Worse, we don't know its value very well. There are efforts underway to measure $G$ more accurately, as reported in Nature earlier this month: It is one of nature’s most fundamental numbers, but humanity still doesn’t have an accurate value for the gravitational constant. And, bafflingly, scientists’ ability to pinpoint G ...


9

You can't calculate the numerical value of Newton's constant from the first principle because it is a dimensionful constant – it has units – so the numerical value depends on the magnitude of the units. And because e.g. the kilogram is defined as the mass of a platinum prototype hosted by a French chateau (the kilogram has the "least objective" definition so ...


1

The gravitational constant can only be determined experimentally. It's empirically derived. There is no physical law we know of that dictates the strength of $G$. Rather, we have the observation that gravitational force is: $F_g = G \frac{m_1 m_2}{r^2}$ Since we can know the other four numbers, we must solve for $G$ experimentally. For a brief background: ...


0

How can an egg be balanced on a nail head on equator? The same way you would balance an egg on a nail head elsewhere. You need to ensure that the nailhead is level, that the egg's centre of gravity is over the nailhead (and ideally, over it's centre) and that the eggs surface at the point of contact is perpendicular to the line from centre-of-gravity ...


0

I am going to assume that you are well aware of and and are comfortable with rotational mechanics of rigid bodies(if you aren't, then you must give the chapter a read at once). I will briefly try to explain it. Consider three points on the ball, the point nearest to the center of the ramp, the center of mass of ball and the point farthest to the center of ...


0

Recently, I have been thinking about alternative causes for the rotation of our planets. You're sixteen. I've noticed that while people of your age can understand Newton's first law of motion, they don't understand the rotational analog of this law. Just as an external force is needed to change an object's momentum, an external torque is needed to ...


2

For the record, here's a worked solution: If $r$ is the distance between the two point masses $m_1$ and $m_2$--which start at rest--then as both accelerate towards each other, where $$ \frac{d^2r}{dt^2} = -\frac{Gm}{r^2} \ \ \ , \ \ \ \ \ m = m_1 + m_2$$ The first step in solving this equation is the least obvious: Multiply both sides by $\displaystyle ...


6

It is great that you "think differently" about problems - that is at the heart of all innovation. When it comes to the rotation of planets, you have to go back to the origins of the solar system: Planets are formed by accretion: a large cloud of debris starts to experience some gravitational pull, and as one "lump" becomes bigger than the others, it starts ...


1

If something is in freefall, starting at $v = 0$ at height $h$, then $$\begin{align} a &= - \frac{GM}{r^{2}}\\ v\,{\dot v} &= - \frac{GM v}{r^2}\\ v\,{\dot v} &= - \frac{GM {\dot r}}{r^2}\\ \frac{1}{2}v^{2} &= \frac{GM}{r} - \frac{GM}{h}\\ v &= \sqrt{2GM\left(\frac{1}{r} - \frac{1}{h}\right)} \end{align}$$ Then, $$\begin{align} J ...


0

The answer depends on the ascent rate $\dot h$. Differentiating gravitational acceleration with respect to time yields $$\frac {d\,g(h(t))}{dt} = \frac {d}{dt}\left(\frac {\mu_E}{(R_E+h(t))^2}\right) = -\frac {2\mu_E}{(R_E+h)^3}\dot h = -\frac{2g(h)}{R_E+h} \dot h$$ where $\mu_E$ is the Earth's standard gravitational parameter, $\mu_E = GM_E$. It's better to ...


0

I'll say no the velocity is constant. At $t=0$ when you throw the ball, you give it a horizontal velocity $\vec{v_0}$. At a certaint time $t$, we have according to the second Newton's law, since the ball is affected only by its gravity as you said (so we neglect any friction):$$\vec{P}=m\vec{a}$$ where $\vec{P}$ is the gravity attraction on the ball and ...


2

I think it refers to the ability of a right handed player/ left handed player to move more in one direction than in other. Nothing to do with physics. Even if it is meant to be scientific its totally wrong .Germans aren't always correct.


0

I'm not sure what he meant by that but he could be referring to an aerodynamic effect. The rotation influences how air flows around it, creating forces, just like lift in a plane. See, for example, Magnus effect: http://en.wikipedia.org/wiki/Magnus_effect Of course, the direction of the curve isn't necessarily 'left' but it depends on the rotation.


0

Your reading is correct, aside from the "Wind Resistance". It is a "drag co-efficient" and is a "fudge factor" so that $A\,C$ is the area of the "effective face" that the falling object presents to the fluid. It measures how much different the effective area is from the actual cross-sectional area. Another way to think of it is as a "tweak" factor to mop up ...


1

We can take three steps or so against a wall, what we do is not walking or running, we kick the wall about one meter (our leg) before the collision, change the rotation of velocity, thus throwing our body upside and taking advantage of friction.


1

A major limitation of dimensional analysis is that you must know some of the physics behind the concept that you are attempting to analyze, and since you have chosen to make the proportionality constant dimensionless, it has skewed your results. However, we can easily deduce the dimensional properties of this constant by rearranging the equation you were ...


0

Also if you have noticed 'c' is coming out to be 1, while it should be -2 . This clearly indicates that constant also has some dimensions. Hence in this case you cannot use dimension analysis to find out the formula. Only way out is the way SIR newton did it.


3

I'll try to explain it like I would to my kid, as soon as he gets there. If you make the same pendulum swing in a horizontal circle, and look at it from the side, you see the same harmonic motion. The only difference is, it stays at the same height all the time, but since we are dealing only with small angles, that's not much of a difference. Now comes, ...


29

In addition to Jim's answer, you could get enough traction if you are allowed to run up the wall with wings (airfoils). Formula 1 racecars could theoretically drive on the walls or ceilings without falling simply because the aerodynamic downforce generated by those wings can be up to 5 times its weight. Of course you'd have to run at superhuman speed in ...


1

Assuming you have overcome the lack of traction, it would be more about the force required than the speed. The required force would also depend on your weight. The force of what we see as gravity is 9.8 N/Kg. Therefore to maintain your position without moving, if you weigh 200Kg, you would need to exert at least 1,960 Newtons; anything more than that and ...


1

You may move up the wall as fast as you want, provided you are able to maintain that speed. The work you need to do in the process is mgR, where 'R' is the radius of the earth, 'm' is your mass and 'g' is the acceleration due to gravity. 'mgR' is the binding energy between you and the earth, which you need to counter to escape earth's pull. If you want to ...


49

Assuming you could get traction against the wall, you could run or walk up it at any speed. However, the problem is that for the large majority of circumstances, you cannot get traction against a vertical wall. The reason we can walk across the ground is because gravity pushes us downwards. This downward force is then opposed by an upward normal force from ...


2

In the absence of angular momentum, then material would be able to follow radial paths to be accreted (because of their mutual gravitational attraction) and thus have a spherically symmetric distribution. For many reasons (see linked questions to the right), the "circum-object" material does have angular momentum, which must be conserved. In the co-rotating ...


7

Does it have something to do with the curvature of the Earth which is assumed to be spherical You'll probably groan when you read this answer since it isn't nearly as complicated as you might think. Essentially, there is factor of $\pi$ since the angular frequency $\omega = 2\pi f = \frac{2\pi}{T}$ A well know result from the linearized pendulum ...


6

First notice that simply by considering the dimension of the parameters involved, one can deduce that the time period of oscillations should go like $$T\propto\sqrt{\frac{\ell}{g}}. $$ This is because $g$ is acceleration hence has the dimensions of Length over Time squared and so the only way the quotient can have the dimension of time is to have the ...


1

You are basically asking how to describe the equations of motion when viewed from an rotating frame of reference (the planets surface). This can be done by including the Coriolis effect. For example if you simplify the problem to just the equator of the planet the equations of motion would then become: $$ \ddot{x}=\frac{F_x}{m}-\frac{v_xv_y}{r}, $$ $$ ...


1

The comments seem to provide the answer. $$ L = p \times r $$ $$ L = m ( v \times r ) $$ Thus if r (the distance to the focus point of the orbit) changes, the velocity can change without the angular momentum changing. This is not possible for circular orbits where v is always perpendicular to r and the magnitude of r is constant. However, in elliptical ...


-1

Today I found this on arxiv.org Shouryya Ray Paper It's the official paper from Shouryya Ray: An analytic solution to the equations of the motion of a point mass with quadratic resistance and generalizations Shouryya Ray · Jochen Frohlich why did he write it two years later from his original solution ?


0

Naturally, I know it to be true that the moon goes around the Earth and that the Earth goes around the sun. While it definitely looks like it if you view it with an astronomy program, it is not entirely correct and a misconception. Sun, earth and moon are not nailed to a specific location, there is no axis stuck in vacuum. Each body attracts each ...


7

No, you don't need to account for the mass of humans. We're a part of the total mass. Children who grow and gain weight are merely transferring mass from other parts of the Earth to them. The mass of humanity versus the mass of everything else on the Earth is a zero sum game. To see whether the Earth is changing mass, you need to look outside the box (or in ...


0

Lots of places state that the Earth's gravity is stronger at the poles than the equator for two reasons: The centrifugal force cancels out the gravity minimally, more so at the equator than at the poles. The poles are closer to the center due to the equatorial bulge, and thus have a stronger gravitational field. TL;DR version: There are ...


5

There is approximately 7 billion people on earth. If we take some average of about 75 kg each, that means $$ m_{\rm population}=7\cdot10^9\times75\,{\rm kg}=5.25\cdot10^{11}\,{\rm kg} $$ The mass of the earth is roughly $$ m_{\rm earth}=5.97\cdot10^{24}\,{\rm kg} $$ which is about 13 orders of magnitude bigger than the total population of the world. So no, ...


6

Yes you need to include the mass of the 7 billion or so humans in the total mass of the Earth, though since their total mass is something like $5 \times 10^{11}$kg and the mass of the Earth is around $6 \times 10^{24}$kg humans make up only about 0.00000000001% of the total mass. We don't make any great contribution to the escape velocity. The point of your ...


2

The Earth/Moon orbit is not truly metastable. As someone alluded to, the moon is actually very very gradually getting further from the Earth. Conversely Mars' moon Phobos is gradually moving closer to Mars over time and Phobos will eventually crash into Mars.


1

From the answer... ..my question I have learnt that Newton's 3rd law of motion is a direct consequence of law of conservation of energy. When a body moves in a certain direction and an opposing force acts on it , it exerts a reacting force (by Newton's law) ... Hence, the body loses its kinetic energy. Problem ....gravitational force: ...


1

My slight issue and where I think I might be missing something, is why doesn't the Earth's speed just mean it can just fly away from the moon and just leave it flying in it's tangential velocity? Even though you have drawn the force of gravity, you are not thinking about it. First of all the force in general is bidirectional, the earth pulls the moon ...


3

You might as well imagine that it is fixed. Basically, as far as the moon is concerned gravitationally, only the Earth exists. For the Earth; our sun. For the Sun; a black hole, ect... and everyone of them has an inherited orbital velocity from their 'parent'. Orbit is not a place. As for why the moon doesn't crash, it formed outside of the Earth's Roche ...


16

Let me try this way: the Sun isn't only pulling on the Earth, it's pulling on the Moon as well. The pull on the Earth is almost the same as the pull on the Moon, so the net effect of the Sun on the relative motion between the Earth and Moon is very small. Recall Galileo's law of motion: if you drop two objects close together from the same height, they ...


3

is why doesn't the Earth's speed just mean it can just fly away from the moon The moon and the Earth fall towards each other due to their mutual gravitation. The Earth doesn't have enough speed, relative to the Moon, to 'just fly away'. Equivalently, the Moon doesn't have enough speed, relative to the Earth to 'just fly away'. Here's an image I ...


0

Here's a simple argument that doesn't require any knowledge of fancy stuff like equipotentials or rotating frames of reference. Imagine that we could gradually spin the earth faster and faster. Eventually it would fly apart. At the moment when it started to fly apart, what would be happening would be that the portions of the earth at the equator would at ...


4

The point is that if we approximate Earth with an oblate ellipsoid, then the surface of Earth is an equipotential surface,$^1$ see e.g. this Phys.SE post. Now, because the polar radius is smaller than the equatorial radius, the density of equipotential surfaces at the poles must be bigger than at the equator. Or equivalently, the field strength$^2$ $g$ ...


1

All celestial bodies lose atmosphere due to a portion of the gas "near space" exceeding escape velocity. The velocity distribution of an ideal gas can be found using the Maxwell-Boltzmann distribution. So an easy approximation for this problem is to say we only want $10^{-6}$ of the molecules to have escape velocity. Using oxygen at 300K, results in an ...


1

If you take something like Neptune and pass it through Earth's orbit perpendicular to the ecliptic so it collides directly with the Earth-Moon system at, lets say, 0.1c you would remove the Earth rather quickly. Get another if you want to disappear the moon as well. Neptune isn't so big that it would disturb everything else, maybe a wobble in Venus or Mars ...


0

Paily's answer is the correct one. Mechanical energy is defined as the sum of kinetic and potential energy and equal to the work done by non-conservative forces. When these are absent (as when they act perpendicular to displacement) they do no work, therefore ΔK+ΔU = 0 and K+U = constant. In short, the principle of conservation of mechanical energy: any ...


-2

You hold a stone and let it go. What work will the Earths gravity do on the stone to bring it down to the ground? This much: $$W=\Delta K=K_2-K_1=K_2$$ Let's remember this for now and do something different. How much potential energy is lost during this fall? This much (energy conservation): $$E_{\text{total }1}=E_{\text{total }2}$$ $$U_1+K_1=U_2+K_2$$ ...


0

Here are a few points to keep in mind: Potential energy is always described as the potential energy of the system. For example, the gravitational potential energy of the Earth-Moon system, belongs to the system as a whole, not the Earth or the Moon individually. So for your example, if you are for instance throwing a brick upwards, it would be the ...


0

If I understand you correctly, your mistake is in using friction as an analog to gravity. Because friction is a non-conservative force the work done is dependent on the path taken. Furthermore the energy "lost" due to friction is stored in a way that is not spontaneously reversible within the system (e.g. heat, plastic deformations, etc.). Gravity on the ...


0

Nothing is actually stored. (You will not find anything "in" the body :) ) The increase of potential energy means in this case that there is a force (of gravity) acting on a body, and the body's movement away from the source of this force increases the distance the body can p o t e n t i a l l y travel under the influence of this force. So if the body is ...



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