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 of the third law will become invalid.
D. for `n` negative, an object ligther than water will sink in water.

Answer

A, C, D

Explanation

`F = (r_(12))/(r_(12)^(13)) GM_(0)^(2) [(m_(1)m_(2))/(M_(0)^(2))]^(n) = (GM_(0)^(2(1 - n)))/(r_(12)^(12))(m_(1)m_(2))^(n)`
Acceleration due to gravity, `g = (F)/(mass) = (GM_(0)^(2(1 - n)))/(r_(12)^(12)) ((m_(1)m_(2))^(n))/(mass)`
Thus acceleration due to gravity on earth is different for different objects. In this situation Kepler's third law will not be valid.
When `n` is negative, then
`g = (GM_(0)^(2(1 + n)))/(r_(12)^(12)) ((m_(1)m_(2))^(-n))/(mass)`
which is positive as `M_(0) gt m_(1)` or `m_(2)`.
Therefore, object ligher than water will sink in water.