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Supposing Newton's Law of gravitation fo...

Supposing Newton's Law of gravitation for gravitation forces `F_1 and F_2` between two masses my and m, at positions `r_1 and r_2` read `F_1=-F_2=-(r_(12))/(r_(12^3))GM_0^2((m_1m_2)/(M_0^2))^(2)`
where `M_0` is a constant of dimension of mass,`r_(12) =r_1 -r_2` and n is a number. In such a case, 

A

the acceleration due to gravity on the 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

Text Solution

Verified by Experts

The correct Answer is:
A, C, D
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.
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