Answer:
U = 5.37*10^33 J
Explanation:
The gravitational potential energy between two bodies is given by:
![U_{1,2}=-G\frac{m_1m_2}{r_{1,2}}](https://tex.z-dn.net/?f=U_%7B1%2C2%7D%3D-G%5Cfrac%7Bm_1m_2%7D%7Br_%7B1%2C2%7D%7D)
G: Cavendish's constant = 6.67*10^-11 m^3/kg.s
For three bodies the total gravitational potential energy is:
![U_{T}=U_{1,2}+U_{1,3}+U_{2,3}\\\\U_{T}=-G[\frac{m_1m_2}{r_{1,2}}+\frac{m_1m_3}{r_{1,3}}+\frac{m_2m_3}{r_{2,3}}]](https://tex.z-dn.net/?f=U_%7BT%7D%3DU_%7B1%2C2%7D%2BU_%7B1%2C3%7D%2BU_%7B2%2C3%7D%5C%5C%5C%5CU_%7BT%7D%3D-G%5B%5Cfrac%7Bm_1m_2%7D%7Br_%7B1%2C2%7D%7D%2B%5Cfrac%7Bm_1m_3%7D%7Br_%7B1%2C3%7D%7D%2B%5Cfrac%7Bm_2m_3%7D%7Br_%7B2%2C3%7D%7D%5D)
BY replacing the values of the parameters for 1->earth, 2->moon and 3->sun you obtain:
![U_{T}=-(6.67*10^{-11}m^3/kg.s)[\frac{(5.98*10^{24}kg)(7.36*10^{22}kg)}{3.84*10^{8}m}+\\\\\frac{(5.98*10^{24}kg)(1.99*10^{30}kg)}{1.496*10^{11}m}+\frac{(7.36*10^{22}kg)(1.99*10^{30}kg)}{1.496*10^{11}m-3.84*10^8m}]\\\\U_{T}=5.37*10^{33}J](https://tex.z-dn.net/?f=U_%7BT%7D%3D-%286.67%2A10%5E%7B-11%7Dm%5E3%2Fkg.s%29%5B%5Cfrac%7B%285.98%2A10%5E%7B24%7Dkg%29%287.36%2A10%5E%7B22%7Dkg%29%7D%7B3.84%2A10%5E%7B8%7Dm%7D%2B%5C%5C%5C%5C%5Cfrac%7B%285.98%2A10%5E%7B24%7Dkg%29%281.99%2A10%5E%7B30%7Dkg%29%7D%7B1.496%2A10%5E%7B11%7Dm%7D%2B%5Cfrac%7B%287.36%2A10%5E%7B22%7Dkg%29%281.99%2A10%5E%7B30%7Dkg%29%7D%7B1.496%2A10%5E%7B11%7Dm-3.84%2A10%5E8m%7D%5D%5C%5C%5C%5CU_%7BT%7D%3D5.37%2A10%5E%7B33%7DJ)
hence, the total gravitational energy is 5.37*10^33 J
Chemical equations show how chemicals interact when a reaction occurs. They use symbols and formulas to show the chemical reaction so that chemical names do not need to be written out. They also help keep track of all elements and the number of atoms of each element on each side of the equation.
Answer:
Hello Todoroki here!
Explanation:
True. the ball may hit any part of the body and still be a legal hit.
<em>Hope this helps! ^^</em>