Calculating the first cosmic velocity / velocity for a circular orbit

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Writing it the way you did, you don't try to compensate the force, but to set it equal. This means you account for the fact that the centripetal force is gravity and nothing more and they do point in the same direction for a circular orbit.

Assume a body in circular motion around the origin of a coordinate system at distance $R$. One can express this motion by: $$ \vec{r}(t) = R \left( \matrix{\cos(\omega t) \\ \sin(\omega t) \\ 0} \right) $$ Therfore we have $$ \vec{v}(t) = \dot{\vec{r}}(t) = R \omega \left( \matrix{-\sin(\omega t) \\ \cos(\omega t) \\ 0} \right) $$ $$ \vec{a}(t) = \ddot{\vec{r}}(t) = - R \omega^2 \left( \matrix{\cos(\omega t) \\ \sin(\omega t) \\ 0} \right) $$

and $ | \vec{r} | = R $, $ | \vec{v} | = R \omega $, $ | \vec{a} | = R \omega^2 $.

The absolute value of force to "allow" this motion is given by: $$ | \vec{F} | = | m \vec{a} | = m | \vec{a} | = m R \omega^2 = \frac{m R^2 \omega^2}{R} = \frac{m |\vec{v}|^2}{R} $$

So if a body is in circular motion the absolute value of force must equal to $\frac{m |\vec{v}|^2}{R}$.

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dimaastronom
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Updated on March 22, 2020

Comments

  • dimaastronom
    dimaastronom over 3 years

    Today in class we calculated first cosmic velocity using this equation $$G\frac{mM}{(R+h)^2}=\frac{mV^2}{R+h},$$ where $m$ = mass of the body, $M$ = mass of the Earth, $R$ = radius of the Earth, $h$ = altitude of the body above the surface of the Earth and $V$= is first cosmic velocity.

    As I understand we need to compensate the attraction force on the left with $ma$, where $a$ is centripetal acceleration. So my question is: "How we can compensate one force with another if both of them are directed to the center of the Earth. Is the equation for compensating force on the right is some kind of imaginary force we assume exists and directed to the opposite side than centripetal acceleration does?".

    • ACuriousMind
      ACuriousMind over 8 years
      What is the "first cosmic velocity"?
    • dimaastronom
      dimaastronom over 8 years
      @ACuriousMind It is the minimal speed needed to get an orbit around the Earth. At least in Russia it called like that ;p
    • Admin
      Admin over 8 years
      Hi no offence of any kind intended, but would you consider altering the title of your post to reduce the confusion of non Russian speaking users. Your post might get more attention that way. Anyway, the very best of luck with your question. Regards
    • dimaastronom
      dimaastronom over 8 years
      @irish physics I did it, thx
    • Admin
      Admin over 8 years
      @dimaastronom Hi I thought about it after and was going to say to you: No leave it, it will catch the user's attention more ...like hmmm "first cosmic velocity" ...wonder what that is, let's have a look...but this a forum based on learning physics, not learning marketing :) best regards
    • image357
      image357 over 8 years
      hey. you changed the title to "escape velocity". this is not what you were looking for! please correct that
  • dimaastronom
    dimaastronom over 8 years
    Ok I thought a little bit more of it and understood that we dont need to compensate the gravity force we just need enough velocity to cover the sufficient distance for not to approach to the Earth but stay on the same altitude. And this is just the second law of Newton. But how we understand that the speed on the right-hand side is the speed we need?
  • image357
    image357 over 8 years
    edited the answer, see above
  • dimaastronom
    dimaastronom over 8 years
    Thx, it helped)
  • Admin
    Admin over 8 years
    @dimaastronom Hi I thought about it after and was going to say to you: No leave it, it will catch the user's attention more ...like hmmm "first cosmic velocity" ...wonder what that is, let's have a look...but this a forum based on learning physics, not learning marketing :) best regards
  • Martin Argerami
    Martin Argerami over 8 years
    @dimaastronom: if you look at the formulas above, the acceleration $\vec a(t)$ is parallel to $\vec r(t)$, i.e. it is radial. So the acceleration is acting on the exact opposite direccion as gravity. This is what justifies going to a scalar equation like the one you posted in your question.
  • image357
    image357 over 8 years
    @MartinArgerami: no this is wrong. The acceleration is antiparallel to the position vector. It shows in the exact same direction as gravity does (inwards).
  • Martin Argerami
    Martin Argerami over 8 years
    @Marcel: you are right. My main point though, is that the direction discussion is missing in the answer, and it is the essential fact that leads to the OP's formula.