Supposing Newton's Law of gravitation for gravitation forces F_1 and F_2between two masses my and m, at positions r_1 and r_2readF_1=-F_2=-(r_(12))/(r_(12^3))GM_0^2((m_1m_2)/(M_0^2))^(2) where M_0is a constant of dimension of mass,r_(12) =r_1 -r_2and n is a number. In such a case,

Supposing Newton's Law of gravitation for gravitation forces F_1 and F

1.	Supposing Newton's Law of gravitation for gravitation forces F_1 and F_2between two masses my and m, at positions r_1 and r_2readF_1=-F_2=-(r_(12))/(r_(12^3))GM_0^2((m_1m_2)/(M_0^2))^(2) where M_0is a constant of dimension of mass,r_(12) =r_1 -r_2and n is a number. In such a case,
Answer» the ACCELERATION DUE to GRAVITY on the earth will be different for different objects none of the three laws of Kepler will be valid only the third law will become invalid for n negative, an object lighter than water will sink in water Solution :Given `F_1=-F_2=-(r_(12))/r_(12)^3GM_0^2((m_1m_2)/M_0^2)^n` `r_(12) =r_1-r_2` Acceleration due to gravity, `g=(\|F\|)/("mass")` `=(GM_0^2(m_1m_2)^n)/(r_(12)^2(M_0)^(2n))xx1/(("mass"))` Here, g is not constant , HENCE constant of proportionality will not be constant in Kepler.s third law . Hence , Kepler.s third law will be invalid But, first two Kepler.s laws will be valid. For negative, n g = `(GM_0^2(m_1m_2)^(n))/(r_(12)^(2)(M_0)^(-2n))xx1/(("mass"))` `=(GM_0^(2(1=1)0)(m_1m_2)^(-n))/(r_12^(2)""(mass))` `g=(GM_0^2)/(r_12^2)((M_0^2)/(m_1m_2))^nxx1/((mass))` Here, `M_0 gt m_1 "or " m_2`. So that `g gt 0` , hence is this case situation will reverse i.e. object lighter than water will sink in water.