Supposing Newton's law of gravitation for gravitation force `F_(1)` and `F_(2)` between two masses `m_(1)` and `m_(2)` at positions `r_(1)` and `r_(2)` read `F_(2)=-F_(2)=(r_(12))/(r_(12)^(3))GM_(0)^(2)((m_(1)m_(2))/(M_(0)^(2)))^(n)` where `M_(0)` is a constant dimension of mass, `r_(12)=r_(1)-r_(2)` and `n` is number. In such a case.

Options

A. the acceleration due to gravity on earth will be different for different objects
B. none of the three laws of Kepler will be valid
C. only the third law will become invalid
D. for n negative, an object lighter than water will sink in water.

Answer

A, C, D

Explanation

`|vec(F)| = (GM_(0)^(2))/(r_(12)^(2)) ((m_(1)m_(2))/(M_(0)^(2)))^(n)`
`= (GM_(0)^(2(1-n))(m_(1)m_(2))^(n))/(r_(12)^(2))` …(i)
`:. g = (|vec(F)|)/("Mass")`
If `m_(1) = M` is considered as mass of earth and `m_(2) = m` as the mass of any other object and `r_(12) = r` is the separation between the earth and the object. Then,
`rArr g = (|vec(F)|)/(m) = (GM_(0)^(2(1-n)))/(r^(2))((Mm)^(n))/(m)`
`= ((GM_(0)^(2(1-n))M^(n))/(r^(2)))m^(n-1)`
`rArr g = ("constant") xx m^(n-1)`
So acceleration due to gravity will be different for different objects (except for n = 1). And Kepler.s third law will not be valid.
For negative n,
`g = ((GM_(0)^(2(1+n))M^(-n))/(r^(2))) = m^(-n+1)`
g will be inversely proportional to the mass m. So, for objects lighter than water, g experienced by the lighter object will be greater than that experienced by the water. Hence, the lighter objects will sink in the water.
The correct option are (a), (c ) and (d).