Hello, guys. So I've been asked by my teacher what exactly is an equation between Potential Energy and Kinetic Energy in gravity law?

I can only solve to:

(G.m1.m2)/r = 1/2(mv^2)

But I want to know the simplified equation, especially to find v. Could you guys help me? Thanks for the help.

The equation for PE is actually the negative of what you wrote:

PE = - \frac {GMm}r

Thus as r gets bigger the value for PE gets bigger as well (actually less negative). So the change in potential energy as an object falls from an initial height of R1_1 to R_2 is:


\Delta PE = -GMm(\frac 1 {R_2} - \frac 1 {R_1})

This change in PE plus the change in KE = 0:

-GMm(\frac 1 {R_2} - \frac 1 {R_1}) + \frac 1 2 m (v_2^2 - v_1^2)=0

You can divide through ny m and rearrange this if you like :

GM(\frac 1 {R_2} - \frac 1 {R_1}) = \frac 1 2 (v_2^2 - v_1^2)

Consider how this equation works for a rock dropped from an infinite height (R_1 = infinity) to the surface of the earth - its impact velocity would be:

v_2 = \sqrt{\frac {2GM}{R_e}} = \sqrt{\frac {2(6.734 \times 10^{-11} \frac {m^3}{Kg-s^2})(5.972 \times 10^{24} Kg)}{6.378 \times 10^6 m}} = 11230 \frac m s

or about 25,000 MPH.

So the equation for escape velocity will be this, right? :

https://www.askmehelpdesk.com/cgi-bin/mimetex.cgi? v_2 = \sqrt{\frac {2GM}{R_e}} = \sqrt{\frac {2(6.734 \times 10^{-11} \frac {m^3}{Kg-s^2})(5.972 \times 10^{24} Kg)}{6.378 \times 10^6 m}} = 11230 \frac m s

It seems my book has a typo while typing this formula (and that's what has keep me busy searching the internet)

Thank you for your help. I really appreciate it.

That's right - escape velocity is the initial velocity needed for an object to be able to "coast" up to an infinite altitude.