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1

Here is one way to solve it. You can multiply both terms by $\dot r$. $$m\frac{d^2 r}{dt^2}\frac{dr}{dt} + \frac{GMm}{r^2}\frac{dr}{dt} = 0 \quad\Longrightarrow\quad \frac{d}{dt}\left[\frac{m}{2}\left(\frac{dr}{dt}\right)^2\right] - \frac{d}{dt}\frac{GMm}{r} = 0$$ Basically $\dot r$ is the integrating factor. We now recognize the potential and kinetic ...

0

Let me summarize the discussion in the comment section. First of all you can safely use Newtons equation of motion $$F = m g$$ with a constant gravitational acceleration $g$. This is simply because the height of a 10 stories building, lets say $30 m$, is still very very small compared to the radius of the earth which is about $6000 km$. Therefore, as ...

4

Yes, $F=ma$, but also $v=at$. That means that, as you fall for a longer time, your speed will increase. After 1 second, you are going at $9.8 m/s$ or $35 km/h$, about the speed of Usain Bolt. After 10 seconds you would reach $98 m/sec$ or $350 km/h$. For a free-falling human, the air resistance actually limits you to about $200 km/h$. When you hit the ...

-2

F=ma is saying that if a mass 'm kg' is accelerated at 'a metres per second per second' then there has to be a constant force of 'F' pushing it. a better equation to represent the effects of gravity would be: F=(G*m1*m2)/d^2 where: G= universal gravitational constant (6.6738410^-11 m^3 kg^(-1) s^(-2)), g= acceleration due to gravity (9.81 m s^(-1) ...

0

When the body is falling down without the affect of air resistance,it will experience only one force,I.e.the gravitational force which pulls out downwards towards the earth..and the object will fall to the surface of the earth will an acceleration 'g'. Coming to the first case...when the object is falling down in the presence of a drag force (friction due ...

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