I am (just for fun) trying to calculate the velocity with which the moon would impact the earth if it stopped orbiting and simply headed towards earth.   I know the force is $\frac{G m_1 m_2}{r^2}$.  How do I calculate a final velocity resulting from a varying acceleration which in turn is dependant on a varying force which in turn is dependant on a varying distance.
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The Problem For a small mass a distance $R_i$ away from the center of the Earth, how long would it take for the object to fall to the surface of the Earth, assuming that the only force acting upon the object is the Earth's gravitational force?  Relevant Information The following discussion seems to have solved exactly the same problem:   However, upon working out the mathematics, I'm not exactly sure how to evaluate the constant of integration.   A Partial Solution $$ F=\frac {-GMm}{s^2} $$ $$ a=\frac {-GM}{s^2} $$ $$ \frac {dv}{dt} = \frac {-GM}{s^2} $$ Multiplying by $v$ and then integrating by $dt$ on both sides, we have $$\frac {1}{2} v^2=\frac {GM}{s} +c_1$$ where $c_1$ is a constant of integration. Substituting initial conditions of $v=0, s=R$, we have $$\frac {1}{2} v^2=GM(\frac {1}{s}-\frac{1}{R})$$  At this point of time, when I use Wolfram Alpha, I get  $$c_2+\sqrt{\frac{2}{R}}t=\frac{\sqrt{s}(s-R)+R\sqrt{R-s}\times{\tan^{-1}(\sqrt{\frac{s}{R-s}})}}{\sqrt{GM(R-s)}}$$  where $c_2$ is a constant of integration. Substituting initial conditions of s=R, t=0, we find that the term $$\tan^{-1}(\sqrt{\frac{s}{R-s}})$$ is undefined. At this point, I'm stuck. Any ideas on where I've made the mistake here?  (For those interested, this question was inspired by the Greek myth which states that a bronze hammer dropped from heaven would take 9 days to hit the Earth and would reach on the tenth).
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When drag force ($bV$) equals to object's weight (mg) then upward and downward force becomes equal. As a result the object comes to rest. If this is true, how is a body moving with constant velocity?
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The force of gravity is constantly being applied to an orbiting object.  And therefore the object is constantly accelerating.  Why doesn't gravity eventually "win" over the object's momentum, like a force such as friction eventually slows down a car that runs out of gas?  I understand (I think) how relativity explains it, but how does Newtonian mechanics explain it?
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I made a naive calculation of the height of Earth's equatorial bulge and found that it should be about 10km.  The true height is about 20km.  My question is: why is there this discrepancy?  The calculation I did was to imagine placing a ball down on the spinning Earth.  Wherever I place it, it shouldn't move.  The gravitational potential per unit mass of the ball is $hg$, with $h$ the height above the pole-to-center distance of the Earth (call that $R$) and $g$ gravitational acceleration.  Gravity wants to pull the ball towards the poles, away from the bulge.  It is balanced by the centrifugal force, which has a potential $-\omega^2R^2\sin^2\theta/2$ per unit mass, with $\omega$ Earth's angular velocity and $\theta$ the angle from the north pole.  This comes taking what in an inertial frame would be the ball's kinetic energy and making it a potential in the accelerating frame.  If the ball doesn't move, this potential must be constant, so  $$U = hg - \frac{(\omega R \sin\theta)^2}{2} = \textrm{const}$$  we might as well let the constant be zero and write  $$h = \frac{(\omega R \sin\theta)^2}{2g}$$  For the Earth,   $$R = 6.4*10^6m$$  $$\omega = \frac{2\pi}{24\ \textrm{hours}}$$  $$g = 9.8\ m/s^2$$  This gives 10.8 km when $\theta = \pi/2$, so the equatorial bulge should be roughly that large.    , the Earth is 42.72 km wider in diameter at the equator than pole-to-pole, meaning the bulge is about twice as large as I expected.  (Wikipedia cites diameter; I estimated radius.)  Where is the extra bulge coming from?  My simple calculation uses $g$ and $R$ as constants, but neither varies more than a percent or so.  It's true the Earth does not have uniform density, but it's not clear to me how this should affect the calculation, so long as the density distribution is still spherically-symmetric (or nearly so).  (Wikipedia also includes , without derivation, that agrees with mine.)
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Is there an equation linking velocity, mass and force? I want to find the maximum force that an object of any given mass could exert. Assuming, the object with the given velocity stops dead in an instant, I thought of using F=ma but there is no time frame for the slowing down of the object since it's instant. Thank you
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So this is the question that's been bothering me:  Say you have a simple rigid body in space that is at rest or traveling with only translational motion at a constant speed. Say that the body is something like a rod and it's not rotating. So, at some point, an external force is acted on the rod, just for an instant and it is not acted along some axis that goes through the center of mass. What will happen??  My guess is that it will start rotating about the center of mass (because of torque) + it will get some translational motion towards the direction of the force. Is that correct? and if so, how much of the force becomes rotation and how much translational movement and why??  Please forgive my English and thnx in advance!!
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I was searching about 'why do not moon crash to earth due to its gravity?'Then by reading physics stake exchange existing questions I came to know that its the sideways motion of the moon that keeps moon in its orbit.   So my question is:- Moon is in circular or ellipthitical orbit and during its orbit moon is always pulled towards the centre of the earth but moon is changing its direction at every point because it is in circular or ellipthitical orbit, so the direction of force pulling moon towards the centre of the earth will also change at every point on the other hand the sideways motion of moon is always in a particular direction then there must be a point in the moon's orbit when sideways motion and gravity will act in same direction and moon should crash to earth.Then why it does not crash?
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How can one analytically calculate the terminal velocity of a settling sphere in 2D? Actually it would be a circular disk. One cannot simple equate boyancy forces minus the drag right? As stated in   the formula $\textbf{U}_s=2/9(\rho_p-\rho_f)\textbf{g}/9\mu$ [1] does not hold!  [1]
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As we know, the velocity of the planet changes as it orbits around the sun, both in its magnitude and direction. The centripetal acceleration causes the change in the direction of velocity but I want to know what is the tangential force that acts on the planet to change the velocity by its $magnitude$. The answer normally as to why the speed changes is: If  the  force  that  the  Sun  exerts  on  the  planet increases  (as  the  planet moves closer),  then  the acceleration   of  the  planet must  increase,  resulting  in a  higher  orbital  speed,  and vice  versa. But this force is not along the body but towards the sun. So such a force should only change its direction.
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Pardon the misleading title.  It is to my understanding that moving/heavy clocks run slow. The Earth itself is under gravitational influence from many sources, and is moving. Is there a way to know how much 'faster' a clock would run if those influences were removed? I don't need a precise number, an order of magnitude is fine, just to get the idea.  Any insight is appreciated.
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If I'm standing at the equator, jump, and land 1 second later, the  Earth does NOT move 1000mph (or .28 miles per second) relative to me,  since my velocity while jumping is also 1000mph.   However, the Earth is moving in a circle (albeit a very large one),  while I, while jumping, am moving in a straight line.   How much do I move relative to my starting point because of this? I  realize it will be a miniscule amount, and not noticeable in practise,  but I'd be interested in the theoretical answer.
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As a layman, i can calculate approx "displacement" just by observing the moving object. And accurately by using a simple "scale". Similarly, again, I can calculate angle from origin by using displacement in $x$ and $y$ dimensions. Similarly I can use a stopwatch and scale to understand velocity.  But when I read about , first of all it confuses with the english word "force", we use in real life. To some extent I am sure, it has nothing to do with that. So exactly what is it in the real sense. How can a layman, see force as, just like he can see displacement or angle? Or force is just a quantity defined by physicists to simplify the combinations of $ma$, they might be facing every time. And thus came up with term "force" ( which is similar in spelling to english word in oxford dictionary). And lastly, why not force has just been called as something proportional to mass and it's displacement or velocity. Why something at the level of change of velocity has been used to define it.
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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.
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Comparing the work required for a mass to escape the Earth's gravity to the necessary initial kinetic energy gives us the escape velocity from the surface of the Earth of around $11\;\mathrm{km\cdot s^{-1}}$.  But assuming no air friction and hence no loss in energy on the axis perpendicular to gravity it shouldn't matter what angle the projectile is projected at.  Is this true?  If there were no air and we fired a projectile at $1^\circ$ to the horizontal at a velocity of $11\;\mathrm{km\cdot s^{-1}}$ should it still escape the Earth's gravity despite such a low ( $11000\sin(01^\circ)\;\mathrm{m\cdot s^{-1}}$) vertical velocity?  Could it be that firing a projectile horizontally at the escape velocity in a frictionless medium would result in a perfect orbit?  Also: Taking escape velocity of $11200\;\mathrm{m\cdot s^{-1}}$ and radius af Earth of $6371000\;\mathrm{m}$ and $g=9.80665\;\mathrm{m\cdot s^{-2}}$. On a flat Earth if you fired an object horizontally at $112000\;\mathrm{m\cdot s^{-1}}$ it would fall  $0.039\;\mathrm{m}$ after $1000\;\mathrm{m}$. But using trigonometry the Earth would have curved away by $0.078\;\mathrm{m}$. Twice the value necessary for an orbit. Can someone check this?
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Newton's 2nd law of motion can't be applied for mass-varying systems. Another force, known as Thrust must come to play. It can be measured using law of conservation of linear momentum. $$\text{Thrust} M.\dfrac{dv}{dt} = v_{rel} .\dfrac{dm}{dt}.$$   where $M$ is the mass of the system;   $\frac{dm}{dt}$ is the rate at which the mass changes;    &amp;    $v_{rel}$ is the relative velocity of the exhausted mass w.r.t to the main system.    This is what my book writes. Now, the left hand side of the eqn. represents force on mass $M$ whereas the right hand side represents the force due to variable mass. How can they be equal? During the application of thrust, $M$ does change and the equation had to be $$ (M - dm). \dfrac{dv}{dt} = v_{rel} .\dfrac{dm}{dt}$$ But that is not the book's equation. Why is mine wrong? What is the intuition?
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Consider some some satellite of mass $m$ around earth, traveling in an elliptical orbit about a focus (earth). Let's suppose at one point $A$ the satellite is a distance $a$ units away the focus at minimum distance, and at another point $B$  it is at a distance of $b$ from the focus, a maximum distance. Let points $A$ and $B$ be collinear with the center of the Earth. When the satellite is at a distance of $a$, it's speed is equal to $v_0$.    Calculate the speed of at the satellite a distance $b$ from the focus.  So first I approached this by the conversion of mechanical energy:  $$\frac{1}{2}mv_0^2-\frac{Gm_e m}{a}=\frac{1}{2}m v_b^2-\frac{G m_e m}{b}$$  I got,  $$v_b=\sqrt{v_0^2+2Gm_e(\frac{1}{b}-\frac{1}{a})}$$  But then I realized conservation of angular momentum gives,  $$mv_bb=mv_0 a$$  $$v_b=\frac{a}{b}v_0$$     I don't see how the two can be equivalent. Is one answer wrong? May someone please explain.
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If I have velocity at perihelion/apphelion, distance away from sun at perihelion/apphelion, and orbital period. How can I find the angular velocity function for earth and subsequently all the other planets in the solar system?