1

I am a math student and I am taking my first physics course of Mechanics. But I don't know the difference between the $w=mg$ force and $F=G\cfrac{m_1 m_2}{r^2}$ (why we don't use the first one when talking about the interaction between our planet and me).

For me, the two forces are related since $g$ is the gravitational acceleration. I know that $F=G\cfrac{m_1 m_2}{r^2}$ is the force with which the earth attracts us (because we have mass).

Qmechanic
  • 201,751
Almhz
  • 15

1 Answers1

0

$w=mg$ is interaction between you and planet only when you are on the surface of the planet i.e when $r=R$ in second equation (Here, $g={\frac{GM_e}{R^2}})$. So, it is a special case.

If you are above the surface of planet at a certain height (say $r$ from the center of the planet), $w=mg$ won't work because now the interaction will $$F=\frac{GM_em}{r^2}\ne{mg}$$

But for small heights interaction can be approximated as $w=mg$.

SarGe
  • 607
  • Still works above/below/anywhere, the sneaky physicist just uses different values for g depending on location. You might have heard that g has different values for different lattitudes. The $9.80m/s^2$ is just an average/approximation. –  May 14 '20 at 03:57
  • Yes. But for simplicity I've assumed value of$g$ to be constant at surface of earth (Ultimately earth to be a perfect sphere). – SarGe May 14 '20 at 04:01
  • Thanks! You helped me a lot! – Almhz May 14 '20 at 04:09
  • And I appreciate all the answers! – Almhz May 14 '20 at 04:09
  • The square root is wrong. It isn’t even dimensionally correct. – G. Smith May 14 '20 at 04:25
  • @G. Smith thanks for your correction. – SarGe May 14 '20 at 04:54