Let us consider two bodies having masses m and m' respectively.
Let they are separated by a distance of r from each other.
As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - 
where G is the gravitational force constant.
From the above we see that F ∝ mm' and 
Let the orbital radius of planet A is 
= r and mass of planet is 
.
Let the mass of central star is m .
Hence the gravitational force for planet A is 
For planet B the orbital radius 
and mass 
Hence the gravitational force 
![f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }](https://tex.z-dn.net/?f=f_%7B2%7D%20%3DG%5Cfrac%7Bm%2A3m_%7B1%7D%20%7D%7B%5B2r_%7B1%7D%5D%20%5E%7B2%7D%20%7D)
Hence the ratio is 
[ ans]
The correct answer as the first one above !
Answer:
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Answer:
Temperature increase = 2.1 [C]
Explanation:
We need to identify the initial data of the problem.
v = velocity of the copper sphere = 40 [m/s]
Cp = heat capacity = 387 [J/kg*C]
The most important data given is the fact that when the shock occurs kinetic energy is transformed into thermal energy, therefore it will have to be:
![E_{k}=Q\\ E_{k}= kinetic energy [J]\\Q=thermal energy [J]\\Re-employment values and equalizing equations\\\\\frac{1}{2} *m*v^{2}=m*C_{p}*dT \\The masses are canceled \\\\dT=\frac{v^{2}}{C_{p} *2} \\dT=2.1 [C]](https://tex.z-dn.net/?f=E_%7Bk%7D%3DQ%5C%5C%20E_%7Bk%7D%3D%20kinetic%20energy%20%5BJ%5D%5C%5CQ%3Dthermal%20energy%20%5BJ%5D%5C%5CRe-employment%20values%20and%20equalizing%20equations%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20%2Am%2Av%5E%7B2%7D%3Dm%2AC_%7Bp%7D%2AdT%20%20%5C%5CThe%20masses%20are%20canceled%20%5C%5C%5C%5CdT%3D%5Cfrac%7Bv%5E%7B2%7D%7D%7BC_%7Bp%7D%20%2A2%7D%20%5C%5CdT%3D2.1%20%5BC%5D)
Answer:
the force of attraction between all masses in the universe; especially the attraction of the earth's mass for bodies near its surface
Explanation:
