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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 `m_(1)` and `m_(2)` at positions `r_(1)` and `r_(2)` read
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)^(3)) GM_(0)^(2) ((m_(1) m_(2))/(M_(0)^(2)))^(n)`
`r_(12) = r_(1) - r_(2)`
Acceleration due to gravity, `g = (|F|)/("mass")`
`= (GM_(0)^(2) (m_(1) m_(2))^(n))/(r_(12)^(2) (M_(0))^(2n)) xx (1)/(("mass"))`
Since, g depends upon position vector, hence it will be different for different objects. As `g` is not constant, hence constant of proportionality will not be constant in Kepler's thrid law. Hence, Kepler's third law will not be valid.
As the force is of central nature. [`:.` force `prop (1)/(r^(2))`]
Hence, first two Kepler's laws will be valid.
For negative n, `g = (GM_(0)^(2) (m_(1)m_(2))^(-n))/(r_(12)^(2) (M_(0))^(-2n)) xx (1)/(("mass"))`
`= (GM_(0)^(2 (1 + n)))/(r_(12)^(2)) ((m_(1) m_(2))^(-n))/(("mass"))`
`g = (GM_(0)^(2))/(r_(12)^(2)) ((M_(0)^(2))/(m_(1)m_(2)))^(n) xx (1)/("mass")`
As `M_(0) gt m_(1) " or " m_(2)`
`g gt 0`, hence in this case situation will reverse i.e., object lighter than water will sink in water
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