a 0.112 kg apple on a tree has a PE of 4.00 J. How high is it above the ground?

Guest Jul 1, 2019

#2**+1 **

a 0.112 kg apple on a tree has a PE of 4.00 J. How high is it above the ground?

**Hello Guest!**

\(F=\frac{G\cdot m\cdot M}{r^2}\\ r=\sqrt{\frac{G\cdot m\cdot M}{F}}\)

F = PE = 4J

m = 0.112 kg

M = \(5.97\cdot 10^{24}\)kg

G = \(\frac{6.672\cdot 10^{-11}m^3}{kg\cdot s^2}\)

\(r=\sqrt{\frac{\frac{6.672\cdot 10^{-11}m^3}{kg\cdot s^2}\cdot 0.112kg\cdot 5.97\cdot 10^{24}kg}{4\frac{kg\cdot m}{s^2}}}\)

\(r=\sqrt{\frac{m^3\cdot kg\cdot kg\cdot s^2}{kg\cdot s^2\cdot kg\cdot m}\times \frac{6.672\cdot 10^{-11}\cdot 0.112\cdot 5.97\cdot 10^{24}}{4}}\)

\(r=\sqrt{9152\cdot 10^{13}\ m^2}\\ r=3339598.05965\ m\)

\(high=3339598.05965\ m-6378 m (Earth's\ radius, geocentric \ mean)\\ \color{blue} high=3333220\ m\)

The tree stands on the south pole and is about 3333221m high.

**Sorry, I read 4 N instead of 4 J.**

\(PE= m\times g\times h\\ h=\frac{PE}{m\times g}\\ h=\frac{4J\times s^2}{0.112kg\times 9,81m}\times\frac{kg\times m^2}{J\times s^2}\\ \color{blue}h=3.641m\)

!

asinus Jul 2, 2019