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]
Answer:
r = √(k q₁ q₂ / F)
Explanation:
F = k q₁ q₂ / r²
Multiply both sides by r²:
F r² = k q₁ q₂
Divide both sides by F:
r² = k q₁ q₂ / F
Take the square root of both sides:
r = √(k q₁ q₂ / F)
Ans. 5 X 10^6
Explanation:
Here,^ represents 6 times 10 and 5 is multiplied. That is 5000 000.
Hope this helps
Thank you for posting your question here. The total distance traveled by the truck during the 3.2 seconds interval is 83 m. Below is the solution:
d = vit + 1/2 at^2
d = (22m/ s) (3.2s) + 1/2 (2.4m/ s^2) (3.2s)^2
d = 83 m
Hope the answer helps.