If I project a body at a very large velocity such that it reaches a height h which is comparable to Earth's radius, then how to derive an equation for 'h' in terms of initial velocity 'u' and other constants.   This is where I ended up with this problem  S = u²/2g  But g is variable with height as h approaches R  So dg = GM/(R+dr)²          --- this is where I think I might be wrong  Hence dS = u²(R+dr)²/GM  So how do I go ahead?
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.
71739
Assume the Earth to be a uniform sphere of mass M and radius R. Also let the Earth be stationary initially. Assume further that there is all other stellar bodies are very far away so as to have no influence in this problem.  A ball of mass $m$ is to be thrown from the surface of the Earth so that it ultimately ends up moving in a circular orbit at a height $h$ from the surface of the Earth. How should the ball be thrown, as in, what should be its initial velocity and, in particular, what should be the angle $\theta$ of the initial velocity with the radial vector from the Earth's center to its initial position on the surface of the Earth?  My attempt at the solution:  Since the Earth-ball system is isolated, its energy would be conserved. Let the initial speed of the ball be $v_0$ and the final speed be $v_f$. We'll have the following equation:  $$ \frac{1}{2}m v_0 ^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2 - \frac{GMm}{R+h}$$   Next, we know that when the ball will finally be in circular orbit, the gravitational force will be providing the necessary centripetal acceleration for it to maintain circular motion, which gives us the following equation:  $$ \frac{mv_f^2}{R+h} = \frac{GMm}{(R+h)^2} $$  Besides, since gravity is a central force, the angular momentum of the ball about the center of the Earth will remain conserved, which yields the following equation:  $$ mv_0R \sin(\theta) = mv_f(R+h) $$  That completes setting up three equations for three variables $- v_0, v_f, \theta $. Solving them, we would obtain the following values for each:  $$ v_o = \sqrt\frac{GM(R+2h)}{R(R+h)} $$   $$ v_f = \sqrt\frac{GM}{R+h} $$  $$ \theta = \sin^{-1} \Biggl( \frac{1+\frac{h}{R}}{\sqrt{1+\frac{2h}{R}}} \Biggr) $$  However, it is observed that for $\theta$, the argument of $\sin^{-1}$ is always greater than or equal to one, being one only if $h$ is zero. The proof may be carried out by initially assuming that the argument of $\sin^{-1}$ is less than or equal to one for $h$ greater than or equal to $0$, and solving the inequality will yield that $h$ must be less than or equal to zero. So the only acceptable solution is for $h$ to be zero. This just winds up being the case of throwing the ball with sufficient speed tangentially on the surface of Earth so that it keeps orbiting on the Earth's surface itself. Thus, there is no possible way to just throw a ball so that it ends up in a circular orbit of certain desired radius.  The problem is that I don't see any physical intuition for why this should be true; either that or I am doing something wrong. Any guidance or help would be greatly appreciated!
49823
For simulation purposes, I'm trying to create a binary star system. If I have a conic equation such as $ax^2 + by^2 = c$, and the masses of two stars, how would I find the $x$- and $y$- velocity vectors (let's assume one star is the reference point, fixed in space), so that one star's orbit is stable and described by the conic equation?
111399
I'm writing this question since I haven't find exhaustive enough answers searching on the Net. I'm writing an application for simulating General Relativity orbits calculating the acceleration from Christoffel's symbols, which are already derived from the Schwarzschild metric in isotropic coordinates:  $$\frac{d^2x^\alpha}{d\tau^2}=-\Gamma{^\alpha}{_\mu}{_\nu} \frac{dx^\nu}{d\tau} \frac{dx^\mu}{d\tau} \hspace{7.1em} (1)$$  My problem is: how should $\frac{dx^0}{d\tau}$ be defined from arbitrary spacial components of the four-velocity?  $$u^i=\frac{dx^i}{d\tau}=\text{arbitrary quantity} \hspace{4.4em} (2)$$  In special relativity, the time component of four-velocity, in terms of the spacial components, is:  $$u^0=\frac{dx^0}{d\tau}=\frac{cdt}{d\tau}=\sqrt{c^2+|u^i|^2} \hspace{4em} (3)$$  where $|u^i|^2 = (u^1)^2+(u^2)^2+(u^3)^2$ is the magnitude squared of the spacial components of the four-velocity. This comes from the Minkowskian metric. Then I derived a similar equation from the Schwarzschild metric in isotropic coordinates:  $$u^0=\frac{cdt}{d\tau}=\sqrt{\frac{B}{A}(c^2+B^2|u^i|^2)} \hspace{4em} (4)$$  where $A=(1-\frac{GM}{2R})^2$ and $B=(1+\frac{GM}{2R})^2$. If this is correct, it's quite straightforward to find that the latter definition gives a greater value than the one from special relativity, since $B &gt; 1$ and $B &gt; A$. Does this mean that the four-velocity magnitude is not constant in General Relativity, as it depends also on radial distance? (I knew that the four-velocity always equals the speed of light)
116863
I am trying to solve the wave equation:  $u_{tt}$  = $u_{xx}$   With initial values:  $u(x,0) =\begin{cases} x^3 - x, &amp;\text{for }|x|\le 1,\ \\0, &amp;\text{for }|x|\ge1\end{cases}$  $u_t(x,0) =\begin{cases} 1 - x^2, \text{for } |x|\le 1, \\ 0, \text{for } |x|\ge1    \end{cases}$  In the domain $D= \{(x,t) | - \infty &lt; x&lt;\infty, \space\space t&gt;0\}  $  I am using D'Alembert's Formula and after inputting the values in I have:  $u(x,t) = \frac{1}{2}[\frac{2}{3}(x+t)^3 + \frac{4}{3}(x-t)^3  - 2(x-t)] $  I am not sure where to go from here.  I think because the initial conditions are not differentiable at some points  then there will be points in the solution that are not differentiable   So I'd need to consider the different cases of $|x+t|\le 1$?  I could be going down the complete wrong path so any help is appreciated.  Thanks
49433
So I have two particles that have collided in 3 dimensional space. I want the particles to rebound off of each other in an elastic manner. How do I determine the resultant velocity vector if I know: Both particles' initial velocity vector, their masses, and their initial position vectors relative to the origin.  I tried to follow the instructions on  but I am not sure how to make their equations work without polar coordinates / how to translate my data to polar coordinates.  Also, if you answer my question in terms of polar coordinates, can you explain which angle is theta and which is phi? Since there are 2 different conventions, I am not really sure which angle is being referred to in the explanation on the aforementioned website.  EDIT: This isn't for a class. I'm writing a physics engine for a gravity simulation I am making using Unity. I haven't taken kinematics since I was in high school, which is why I am so rusty.
5963
Suppose you have an object some distance from you and moving at a velocity different to the Hubble velocity you'd expect at that point. How does the motion of this object change with time? Does it travel in a straight line (i.e. geodesic), and if you wait long enough how far will it get?
103176
So I was thinking about this really simple situation and then I came across a little question. The situation is as follows: There is a block of wood suspended from a ceiling like support with two strings. The strings are perfectly rigid and light with no friction between various parts of the string. .A bullet is shot horizontally and gets embedded in the block My question why can we conserve momentum in the horizontal direction in this scenario when there is a component of tension acting in the horizontal direction which is an external force force. Why do we consider that the bullet block system has a horizontal velocity at max height? I get that $\frac{(M+m)v^2}{r}=(M+m)g$ And $\vec{F_{net}}=0$ at this point but there is still a velocity component perpendicular to the direction of the string which means the system will rotate about point of suspension further and has a greater height achievable which is a contradiction In conclusion shouldn't the equation for max height be $1/2mv^2=(M+m)gh_{max}$ rather than $1/2\mu v^2=(M+m)gh_{max}$ where $\mu$ is the reduced mass of the system Please correct my conceptual misunderstandings and shed some light on situation. Thanks in advance.
1998
I'm calculating the force of a body hitting water from a high height. So far I've figured out the velocity at the time of collisions would be 40 m/s. The body is 78 kg. I seem to need the time it takes for it to accelerate to a stop after hitting the water but I can't find any concrete numbers for that, what would be a good estimation?
3962
I am preparing for my quals and stumbled across the following problem, and although it only requires undergraduate-level physics, I feel I can't piece everything together.  "A rocket of mass $m_0$ is propelled by a giant monochromatic laser mounted on the back of the rocket. The laser emits a beam with a power of $P_0$ watts and a frequency $f_0$, both measured in the rest frame of the rocket. When the beam is turned on, the rocket is driven in the opposite direction by the recoil.  (a) At $t=0$, the laser is turned on, with the rocket's velocity initially at rest in the Earth's reference frame. Calculate the instantaneous acceleration of the rocket.  (b) If the rocket is moving at velocity $v$, what is the instantaneous beam power as measured in the Earth's reference frame?   (c) The laser is kept going until the speed of the rocket reaches $v=0.9c$. What is the rest mass of the rocket at this point?"  I work in units where $c=1$. For (a), in the Earth's reference frame, I get a momentum $p(t) = \gamma m_0 v + h f$, where $\gamma = \frac{1}{\sqrt{1-v^2}}$ and $f = f_0 \sqrt{\frac{1-v}{1+v}}$ because of the Doppler shift in frequency as the rocket travels further away. Differentiating w.r.t time, I get  $$ \frac{dp}{dt} = \gamma^3 m_0 a - \frac{\gamma a}{(1+v)} h f_0 $$  This is as far as I get. Equaling $dp/dt=0$ for conservation of momentum make the $a$ cancel out, and there isn't any other time dependency to differentiate $v(t)$.  For part (b), we can say that in the frame of the rocket, $P_0 = r h f_0$, where $r$ is the rate of emission of photons. In the frame of the Earth, we can therefore write   $$ P(t) = \frac{dE}{dt} = \frac{d}{dt} \left( \gamma m_0 + r h f \right) = m_0 \gamma^3 v a - \frac{\gamma a}{(1+v)}  P_0 $$  but since I don't have $a$, there's not much I can do. Also, I am not sure if $r$ needs to be corrected with a proper time factor when we boost between frames.  Also, for part (c), my first intuition was to say that the rest mass was $m_0$, but now I am thinking that the rocket's total mass decreases because we need to account for the energy lost from the laser. I am slightly confused as to what I need to take into account.
117465
Suppose we have an equation of  motion as $$\frac{d^2x}{dt^2} = -kx + c,$$ then can it be called a SHM? Since acceleration is still proportional to displacement. But then, how will we define the mean position?
106534
So I would have thought that this would be how you derive the  on a spring: basically the same way you do with gravity and other contexts, use $$W=\vec{F}\cdot \vec{x}.$$ If you displace a spring by $x$, then it exerts a force $-k x$, so $F=-kx$, since the displacement is $x$.   So $$W=-kx^2.\qquad \leftarrow\text{ (however, apparently wrong!)}$$  I've seen the correct derivation of work in a spring (with an extra half) and don't doubt that it's correct, but also don't see where my logic fails in this alternate derivation.
16367
This is a question from my Calculus book.  I was showed the correct answer but I really need to learn what the process is to achieve the answer.  Here is the question:  If a ball is thrown into the air with a velocity of $40 \frac{ft}{s}$, its height in feet $t$ seconds later is given by $y=40t - 16t^2$ .  Find the average velocity for the time period beginning when $t = 2$ and lasting   (i) 0.5sec   (ii) 0.1sec   (iii) ...
70429
In the derivation of drift velocity I have seen two variations and want to know which one's correct.   $s=ut+\frac{at^2}{2}$ Assume that the drift velocity of any electron in any conductor is : $$\vec{v_d}=\frac{\vec{l}}{t}$$ Due to the electric field the acceleration of electrons in any conductor is: $$a=\frac{-e\vec{E}}{m}$$ Now the distance travelled by an electron after a long time (initial thermal velocity = 0) $$l=\frac{at^2}{2}\implies \frac{-e\vec{E}t^2}{2m}$$ the time between the collisions is $\tau$ $\therefore$ $$l=\frac{-e\vec{E}\tau^2}{2m}$$ thus the velocity is $$\vec{v_d}=\frac{-e\vec{E}\tau}{2m}$$ In another proof I saw the author using $v=u+at \implies v_d=\frac{-e\vec{E}\tau}{m}$   My question is which of the the two equations of motion can be used in the proof? Can they be used at all.
90395
We know that if an object has been lifted a distance $h$ from the ground then it has a potential energy change:  $$\Delta U = mgh $$  so $h$ is proportional to $\Delta U$.  However, we have also the gravitational potential energy law:  $$ U= -\frac{G M m}{r} $$  where the distance is inversely proportional to the potential energy.  What did I miss? Is the distance of the object proportional or inversely proportional to the potential energy?
42940
I need to calculate the initial velocity required to launch a projectile at a given angle from point A to point B. The only force acting on the projectile after launch will be gravity – zero air resistance. The projectile is launched within a simulated, virtual environment; however, I am asking for help with the physics rather than the simulation itself.  I have had success (with a different equation) when point A and point B are at the same height; however, once point B is at a different height, my calculations become less precise – well, wrong in fact.  I researched the following formula for finding the range of a projectile on uneven ground. The parameters are all available to me, except the initial velocity that I need to solve for.     $d$ : range or distance   $v_i$ : initial velocity   $g$ : gravity   $\theta$ : launch angle   $y_0$ : launch height   $$d = \frac{v_i \cos\theta}{g}(v{_i} \sin \theta + \sqrt{(v_i \sin\theta)^2 + 2gy_0})$$  I attempted to solve for initial velocity (eq: A):  $$v_i = \sqrt{\frac{d^2g}{2\cos\theta^2(y_0+d \tan\theta)}}$$    Using this equation in my simulation I apply the velocity to a normalised displacement vector and launch the projectile. It gets close to its target but:   $x_{final}$ is always correct $z_{final}$ is always incorrect - it is close to the desired $z$ but always offset by a seemingly proportional amount.    I have spent many hours trying to review the equation but I have been unsuccessful. Any advice would be most appreciated:   is this the correct equation? have I solved for $v_i$ correctly? I have looked for an example equation online but I have not found one structured how I need it, i.e. solving for $v_i$ I have spent hours researching online - perhaps I have missed some good (but entry-level) resources. Do you know of any reference material that may help me?   Update - I have now named the equation above for referencing below  Following on from the answer by @Pygmalion, which I am still gratefully working to understand:  I agree that your derived equation is equivalent to mine (A). Using yours (@Pygmalion's) in my simulation, I therefore observe the same failures: the projectile always lands short of the target.  I have simplified the simulation keeping the launch height and target height the same. I still encounter the same problem with equation A; however, when passing the same parameters to the following equation the projectile always hits the target precisely (eq: B):  $$v_i = \sqrt{\frac{dg}{\sin2\theta}}$$  Given the success of equation B when the launch and target heights are the same and the failure of equation A given the same conditions, I question whether equation A is the correct one to solve the problem since I would expect it to work for any combination of launch|target height.  I think the help I need is around why equation B works. Does it cater for conditions that  equation A overlooks? If equation A is unsuitable, are you able to recommend an alternative?  In short, the $v_i$ calculated by equation A (in my simulations) is always less than $v_i$ from equation B given the same inputs - the projectile therefore always falls short of its target.
117101
I'm trying to solve the two-body problem numerically, setting up G, m1 and m2 =1.0.  The masses are placed at the positions -10 and 10 respectively along the x-axis and gave them both 0 on the y-axis. I am having some real issue with the initial conditions fitting for a keplerian orbit with a given eccentricity (e=0.9)! Also how would the initial condition change for different mass ratios?