Usually when on Earth we can use as an approximation:  g = a = 9.18 $\frac{m}{s^2}$ s = vi + 1/2 a$t^{2}$  However, how would I determine the distance traveled in situations where the the acceleration due to gravity is changing non-negligibly because the distances are so large? I know that the following equation has to be used somehow:  a = $Gmr^{-2}$  I am not sure exactly how though. For example, if given the information that an object is at a certain distance and is being attracted by a larger object of a given mass, the method should calculate where the object will be at a given time.
59620
I thought this would be a simple question, but I'm having trouble figuring it out. Not a homework assignment btw. I am a physics student and am just genuinely interested in physics problems involving math, which would be all of them.  So lets say we drop an object from a height, $R+r$, it falls toward earth. This height, $R+r$, is far enough away that the $g$ it experiences is a fraction of $g$ at sea-level. Let's just say that air resistance is negligible, and it wouldn't be that complicated to just integrate from $0$ velocity to the terminal velocity piece wise and deal with the rest of it later.  So the key here is that acceleration is changing with time. I thought I could simplify this by saying it changes with distance, and it has nothing to do with time, but this didn't really help, my guess is that maybe time is important (doh).  I tried integrating acceleration with time, and ended up nowhere. I tried integrating $a=GM/R^2$ with respect to $R$ from $R+r$ to $R$ and ended up with a negative function.  I saw somewhere someone tried to expand with taylor series, they even have something similar on hyperphysics, but I can't figure out how to obtain the polynomials that precede the variables.    This is the hyperphysics site where they use polynomials to find the distance.   Maybe I can't solve this because I haven't taken a course in differential equations yet. What I want to know is how to calculate the distance at any time.
66800
There is a problem with my logic and I cannot seem to point out where. There's a rocket ship travelling at close-to-c speed v without any acceleration (hypothetically), and there is an observer AA with a clock A on Earth, and there's another observer on the rocket BB with a clock B and these two clocks were initially in sync when the rocket was at rest with a FoR (frame of reference) attached to Earth. Now, this rocket's moving and AA tells us that B is running slower than A by a factor of$$ \gamma = 1 /(1-v^2/c^2)^{1/2} $$ $$ t_A/t_B = \gamma $$ where v is the relative 1d- velocity between the two i.e the earth and the rocket! That would mean the time elapsed on A is greater than that on B but this will happen only in the FoR of AA? So $t_B$ in this equation must be the time on B as observed by AA? Is this correct? What do the terms mean in the equations? If the symmetry holds and BB doesn't accelerate, then BB could say that $$ t_B/t_A = \gamma $$ right? where $t_B$ and $t_A$ are the times on B and A with respect to FoR of BB? But I was solving this problem and I took the earth FoR of A, but the prof took the rocket FoR of B? Like how will I know which FoR to solve the problem from? It'd greatly help if the terms in all the above equations were laid down neatly! DO we even need these FoRs?? Because in all the solved problems the prof is't specifying any and is using random ones! Please help!!!! This is the question where I messed up. The first rocket bound for Alpha Centauri leaves Earth at a velocity (3/5)c. To commemorate the ten year anniversary of the launch, the nations of Earth hold a grand celebration in which they shoot a powerful laser, shaped like a peace sign, toward the ship.  According to Earth clocks, how long after the launch (of the rocket) does the rocket crew first see the celebratory laser light?  This must be 25 years. My reasoning is: If v = 3c/5 10v + vt = ct where t is time taken by the light to reach the rocket from Earth as calculated from Earth.. and I solved that for t. and added 10 years to that because the time starts at the launch of the rocket!  According to clocks on the rocket, how long after the launch does the rocket crew first see the celebratory laser light?  This is 20 years. Here, I say: If it takes 25 years as observed by clocks on earth for the laser to reach the rocket, what should be the corresponding time as seen on a clock on the rocket? Using the formula: 25 = $\gamma$t where $\gamma$= 5/4 solved for t!  According to the rocket crew, how many years had elapsed on the rocket's clocks when the nations of Earth held the celebration? That is, based on the rocket crews' post-processing to determine when the events responsible for their observations took place, how many years have passed on the rocket's clocks when the nations of Earth hold the celebration?  For this, I did the following: 10 years on earth = T years on rocket ship where T must be lesser than 10 as observed from Earth FoR! Therefore, T= 4(10)/5 years = 8 years! But, prof says, 10 years on Earth = T years in rocket ship where T must be GREATER than 10 as observed from the rocket FoR??? Therefore, T = 10(5/4) years = 12.5 years!! What does this question actually want?
22562
In the Wikipedia article on  I see two formulae for the gravitational time dilation. One contains $\sqrt{1 - r_0/r}$ but the other $\sqrt{1 - 1.5 r_0/r}$. What formula should I use to calculate the time dilation due to gravity inside a satellite? The speed induced time dilation is of course just special relativity.  With the first of these two formulas I can perfectly reproduce the graphs on time dilation as a function of $r$,  I find that the two contributions to time dilation cancel at 1.5 Earth radii, all numbers are OK... But why not the second formula?
57940
We know a mass of an object of one kilogram as an object that weighs W = mg = 9.8 N and we reference it to that, (when it should as a fundamental parameter describe weight not the opposite). But if we were to describe mass to an alien civilization on an alien planet we are exchanging knowledge with, by sending them a one kilogram object, according to their gravity they will measure it differently.  Also their star could be curving space time in such a way, or their velocity according to SR will cause them to perceive mass on our planet -if they observe from away- differently.  If we asked a crew on a space ship moving at a speed close to the speed of light wrt us, or moving in a gravitational field they don't know about, to measure the mass of our planet, they will get different results. On the same principle, we could be measuring the mass of far celestial objects like planets, differently.   I perceive space time as full of curves and irregularities. We know the about some of these and we don't about others. I mean we know about the earth's gravity, the sun's but we ignore the effect of the galaxy and the cluster, dark matter, and who knows what else. Besides we change our position all the time with relation to these external factors that bend spacetime. So our measurements to an outside neutral observer, say these aliens could change.  I've found definitions of  like "the quantity of matter in an object" but that seems like the good old: mass = volume x density, but mass is the more fundamental quantity than either of them, with which theses parameters should be described by mass not the opposite, not to mention how relative these other quantities are, considering SR and GR. Or "The resistance of an object to acceleration" but again you have to describe how fast that object is moving and what spacetime it's in according to SR and GR.  So what makes us so confident that mass is such a universal value, when we built everything on a concept referenced by our own gravity, and maybe our own reference frame?  How do we describe mass to the aliens, who don't know about our (g)?  How do we measure the mass of celestial objects say planets, by units like kilograms, and pounds, while they are not subject to the earth's gravity, (I'm well aware of the difference between weight and mass). I mean that's what they will weigh -ignoring their own gravity- if they were in our atmosphere on a huge scale? So what's the method?  Bottom line Is there a direct way to   like we do with other fundamental values like length and temperature, other than using a scale or , i.e not depending on other physical parameters to describe it?  --Forgive the length..
58552
In classical General Relativity (meaning not modified) one can think of geodesics in two ways.    One way is to say that a geodesic is the curve which is the straightest (in analogy with the flat case) among all curves. More or less the story goes like this (correct me if I'm wrong): in the flat case geodesics are of the form $x^\mu(s)=st^\mu +b^\mu$ where $t$ and $b$ are constant vectors and $s$ is the curve parameter. The curve tangent vector is $\frac{dx^\mu}{ds}=t=const$ so $$\frac{dt^\mu}{ds}=t^\nu\partial_\nu t^\mu=0.$$ This is a tensor equation and by general covariance it  holds true in a general spacetime, mutatis mutandis: $$t^\nu\nabla_\nu t^\mu=0.$$ Another way is to obtain the equation by finding the minimum of the length functional $$L=\int \sqrt{g(t,t)} \,\, d\tau,$$ we get $$\frac{dt^\mu}{d\tau}+\Gamma^\mu_{\alpha\beta}t^\alpha t^\beta=0.$$ The above equation turns out to be just $t^\nu\nabla_\nu t^\mu=0$, with a certain parametrization choice.    The two formulations are thus equivalent. My question is when and why is it so? Is there a deep reason?  Looking up Wald's book I found the following, unclear to me, argument:     "On a manifold with a Riemannian metric, one can always find curves of arbitrarily long length connecting any two points. However, the length will be bounded from below, and the curve of shortest length connecting two points (assuming the lower bound in length is attained) is necessarily an extremum of length and thus a geodesic. Thus, the shortest path between two points is always a straightest possible path."    Later in the book he also says something about conjugate points.  I always thought it had to do with torsion: If we relax the condition $\Gamma^\mu_{\alpha\beta}-\Gamma^\mu_{\beta\alpha}=0$ then the connection is not Levi-Civita, hence $\delta_g L=0$ remains unchanged but $t^\nu\nabla_\nu t^\mu=0$ does and gives a different geodesic equation.. so Straightest $\neq$ Shortest anymore.  Can anyone clarify things to me?
99658
I remember the first time I saw this formula, $F = ma$, and it seemed perfectly correct and natural to me (in the sense that it is very straight-forward and logical, not that it is easy by any means). But then, I wondered, what if an object is moving with a constant speed? Maybe a car. If it travels with constant speed and hits me, it still hurts. So it's force must not be $0$.  One more thing related to this problem. If the same car is pulling another car behind it, if the car goes with constant speed, it implies that the first car doesn't exert any force to the second car at that moment, since it is not technically pulling it anymore (espacially if we don't take into consideration the friction).  So my question would be the following: How do these two phenomenons describe the same law of motion?   PS: If gravity wasn't behaving as some kind of acceleration pulling me constantly to the Earth's core, then if I jumped upwards, would I start slowly getting away from Earth?
90598
Greetings StackExchange!  We were having a conversation with a peer about stupid ways of interpreting theories. We would go to and fro with an interpretations, but we would always find a way to disprove them. There was one, however, that we couldn't refute:  Gravity on a plane in a bidimensional space can be interpreted as the acceleration of spacetime towards an object.   We tried to invalidate in saying that the inertial mass and gravitational mass would be different, thus violating GR, but after some thought this couldn't be. The gravitational mass would accelerate spacetime towards it. If that object we were to be pushed, the pulling of spacetime would make it difficult for it to move.   We know for a fact this can't be true, and we would greatly appreciate any refutation of this thought. I can clarify if needed, or do a sketch. I apologise in advance as I'm not a native speaker.
65341
I would like to recover the (timelike) geodesics equations via the variational principle of the following action:  $$ \mathcal{S}[x] = -m \int d\tau = -m \int \sqrt{-g_{\mu\nu}\,dx^{\mu}\,dx^{\nu}} $$  Using an arbitrary auxiliary parameter $\lambda$, then one is able to rewrite the action and obtains: $$ d\tau = \sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$ $$ -m \int d\tau = -m \int (d\tau / d\lambda) d\lambda =  -m\int\sqrt{-g_{\mu\nu}\, \tfrac{dx^{\mu}}{d\lambda}\tfrac{dx^{\mu}}{d\lambda}} d\lambda $$  Now we can vary the paths $x^{\mu}(\tau) \rightarrow x^{\mu}(\tau) + \delta x^{\mu}(\tau)$.  I need the following equation to be true, so i get the geodesics equations, but I don't know how one can say the following: $$ \int d\tau \,\delta g_{\mu\nu}\frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} = \int d\tau\, g_{\mu\nu, \rho} \frac{dx^{\mu}}{d\tau} \frac{dx^{\nu}}{d\tau} \delta x^{\rho} $$ I know there is a simpler action, which leads to the same equation of motion, but there one has to do the same manipulation, which I sadly don't understand.  Thanks for helping me!
12154
As far as I’m aware, gravity in general relativity arises from the curvature of spacetime and is equivalent to an accelerated reference frame. Objects accelerating in a gravitational field are in fact inertial and are moving through geodesics in spacetime.  So it could be said then that it is not really a force, but a pseudoforce much like the Coriolis effect. If so, why is it necessary to quantise gravity with a gauge boson, the graviton? And why is it necessary to unify it with the other forces?
70006
We all know $a = \frac{\mathrm{d}v}{\mathrm{d}t}$. A little application of the chain rule leads to the relation $$a=v\frac{\mathrm{d}v}{\mathrm{d}x}$$ But the above equation shows that $a=0$ whenever $v=0$. And this must be wrong, as when we throw something vertically from the earth surface it stops and then returns due to gravity. There is always an acceleration in the downward direction.  I have come up with two possible ways to solve this problem:   We know that $v = \frac{\mathrm{d}x}{\mathrm{d}t}$, so whenever $v=0$ then $\mathrm{d}x=0$ (i.e. no displacement in that infinitely small time). Also, $$a = \frac{\mathrm{d}v}{\mathrm{d}t} = \frac{\mathrm{d}v}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}$$ only if $\mathrm{d}x$ is not equal to $0$, as multiplying both numerator and denominator with $0$ will make it $0/0$ (undefined). Or multiplying by $0/0$ is not equivalent to multiplying by $1$.  Now since $\mathrm{d}x=0$ whenever $v=0$, therefore we can not write $a = v\frac{\mathrm{d}v}{\mathrm{d}x}$ when $v=0$. If we plot a $v$-$x$ curve for the motion, whenever $v=0$, as explained, $\mathrm{d}x=0$. Therefore, at that instant $\frac{\mathrm{d}v}{\mathrm{d}x}$ (the slope of the $v$-$x$ curve) will be not defined (as $\mathrm{d}x=0$), therefore the equation $a = v\frac{\mathrm{d}v}{\mathrm{d}x}$ will be undefined and we cannot determine acceleration at that instant with the $v$-$x$ curve.   Now that actually means that by plotting $v$-$x$ curve we lose information about acceleration of the particle when $v=0$.   I want a method to find the acceleration at that point using $v$-$x$ curve. Also a explanation for this defect of the $v$-$x$ curve.  Also I would like to add that so far no physics book (that I had read) has explained this before writing this relation. Also they do not mention that this will not work for $v=0$.     I would like to add the situation where I observed this: when you throw something vertically upward with velocity such that it reaches to a height 10 then the v-x curve will be-   the slope of above curve is clearly not defined at x=10(or the instant where velocity is 0) But we know that there was a constant acceleration throughout the flight. Then how can we find acceleration at that instant??? Also assume that we only have v-x curve. Or if we cannot then why??
71810
Problem:  $F = \frac{(G m_1 m_2)}{r^2}$  An object of unknown mass is placed directly between the Earth and the Moon such that the forces of gravity acting on the object are equal from both the Earth and the Moon. The distance from the Moon to the object is exactly one-tenth of the distance from the Moon to Earth. Using the above formula, determine the distance from the Moon to the Earth.  Given is:  The factor $G≈6.67∙10−11\mathbf{N(\frac{m^2}{kg^2})}$ is a universal constant.  Mass of Earth $≈5.972∙10^{24}\mathbf{kg}$   Mass of the Moon as $≈7.348∙10^{22}\mathbf{kg}$  Mass of unknown object = $m$  My solution:  $F = \frac{(6.67 *10^-11) m (7.348*10^{24})}{R^2}$  $F = \frac{(6.67 *10^-11) (5.92* 10^22) m}{10R^2}$  As gravitational force acting on the object is equal   $\frac{(5.92\cdot 10^22)}{100r^2} = \frac{(7.348*10^{24})}{r^2}$  $R = 1.24 \cdot 10^4 KM$  Is this a viable solution to the problem? Please provide feedback. Thank you!
23500
Related to this question:   $E=mc^2$ - energy equals mass. So, if an object has gravitational potential energy relative to another object, it should have additional mass to represent that extra energy.  So, if I have an apple in my hand, it should have quite a large amount of gravitational potential energy relative to a black hole at the edge of the observable universe. I did a quick calculation of how much this gravitational potential energy would be (using Newton's Law of Gravitation) and it came out to be the equivalent of around 500 tons (and that's just for one black hole!).  So, why does my apple not weigh a lot more than it appears to?  EDIT: Here's my calculation of the GPE:  Assumptions:   The apple has mass $m_a=0.3 kg$ The black hole has a mass equal to 1000 solar masses, $m_b=2\times10^{33} kg$ Distance between the apple and black hole is 60 billion light years   Obviously $GPE=mgh$ can't be used, because the gravitational acceleration is not constant and varies with distance, so it has to be integrated.  I used Newton's Law of Gravitation to provide the acceleration:  $$ F=\frac{G{m_1}{m_2}}{h^2} $$  A small change in height produces a small change in energy, which can be used to integrate:  $$ \begin{eqnarray*} \delta E&amp;=&amp;F \delta h\\ \int_0^EdE&amp;=&amp;G{m_a}{m_b}\int_1^R\frac{1}{h^2}dh \end{eqnarray*} $$  I'm integrating the force starting at 1m from the center of the black hole, as the force goes to infinity at $h=0$. It's a little crude, but it should provide a conservative estimate.  $$ \begin{eqnarray*} E&amp;=&amp;G{m_a}{m_b}(1-\frac{1}{R})\\ E&amp;\approx&amp; G{m_a}{m_b}\;\;\;\;\;\;(R&gt;&gt;1)\\ G&amp;=&amp;6.673\times 10^{-11}\;\;N(\frac{m}{kg})^2\\ G{m_a}{m_b}&amp;=&amp;4\times 10^{22}\;\;Nm\\ E&amp;=&amp;mc^2\\ \Delta m&amp;=&amp;\frac{4\times 10^{22}}{(3\times 10^8)^2}\\ &amp;\approx&amp;444,000\;\;kg \end{eqnarray*} $$  So, a little short of 500 tons, but still pretty big!
121125
Displacement in an accelerated classical object is: $$s=ut+\frac {at^2}{2}$$ What is the displacement of an accelerated relativistic object?  In Newtonian mechanics there are two types of displacement.   Displacement of an object with velocity constant: $$s=ut$$ Displacement of an accelerated object with acceleration constant: $$s=ut+\frac {1}{2}at^2$$   This is not completely clear, but I think the second displacement should be something different in relativity. Is that true?  What is the result of this integral in relativity? $$s=\int (u+at)dt$$ classicaly $$s=\int (u+at)dt=\int udt +a\int tdt=ut+a\frac {t^2}{2}+x_0$$  What is the relativistic one?  more details: $$v=u+at$$ $$s\to displacement$$ $$a\to acceleration$$ $$u\to initial \,velocity$$ $$v\to final \,velocity$$
12210
Newton's second law of motion states that $f = ma$. However, in this equation, theoretically there could be a value of $f$ and $m$ that results in an acceleration that is enough to push an object past the speed of light. In my case I have a value of newtons that is large enough (I think) to accelerate an object past the speed of light if $f=ma$ held. Is there any way, knowing that value of newtons, and the mass of the object it is acting on to find the acceleration in an Einsteinian universe?
104656
So, according to Einstein's theory of general relativity, gravity is not a force instead it is a consequence of objects with mass deforming spacetime, right? And so, according to him, there is no difference between standing on a platform accelerating upwards at 9.81 m/s^2 and standing on Earth, right? But, there is a difference. At least, I think it's a valid difference. If you drop two apples from a high altitude on Earth, not only will they accelerate downwards at 9.81m/s^2 but they will also slightly move towards each other because the Earth is a sphere and gravity acts radially inwards pulling the two apples closer to each other. But if you drop the apples from the same height above the accelerating platform, the apples will only appear to accelerate downwards at 9.81 m/s^2 but will not move towards each other like they would on Earth. Why is this true? Is this even a valid difference? If it isn't, why isn't it?
32545
I know this may seem a weird question, but it always bothers me. My physics book (Resnick,Halliday,Walker), and also various sites never say anything beyond acceleration.   But when a moving body is being acted by a variable force , its acceleration will definitely change: it will either increase or decrease. Then there will be rate of change of acceleration with respect to time. So, why don't books mention this? What is the cause for not measuring $\frac{d\vec{a}}{dt}$ ? If it exists, what is the use of it?
82924
How much time would it take for a traveler (in traveler's perspective) to reach a star at distance $d$, if it accelerates at 9.8 m/s² (for a comfortable travel)?  I understand that $V = V_0 + a\times t$ and such related high-school physics, and I know there is something called Lorentz transformations that I can grab the formulae from the Internet, but I can't really put the numbers together to get what I want, so I would also like to know how to derive the answer to the question from the basic equations.  Also, it is not quite clear to me the role of the $E = mc^2$ in this travel, specifically: in travellers perspective, can it accelerates at 9.8 m/s² indefinitely, using always a constant energy output? In other words, would the ship's engine "feel" the ship heavier, or the mass increase effect would only be observable from another frame of reference?