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Nokitron · Aug 28, 2013, 05:20 AM

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.

ebaines · Aug 28, 2013, 05:52 AM

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:

$<br /> \Delta PE = -GMm(\frac 1 {R_2} - \frac 1 {R_1})$

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.

Nokitron · Aug 28, 2013, 03:25 PM

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

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

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.

ebaines · Aug 29, 2013, 05:34 AM

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