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Question
Supposing Newton’s law of gravitation for gravitation forces F1 and F2 between two masses m1 and m2 at positions r1 and r2 read F1 = – F2 = `- r_12/r_12^3 GM_0^2 ((m_1m_2)/M_0^2)^n` where M0 is a constant of dimension of mass r12 = r1 – r2 and n is a number. in such a case.
- the acceleration due to gravity on 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
a, c and d
Explanation:
Given, `F_1 = - F_2 = (-r_12)/r_12^3 GM_0^2 ((m_1m_2)/m_0^2)^n`
Acceleration due to gravity, `g = (|F|)/"mass"`
= `(GM_0^2 (m_1m_2)^n)/(r_12^2 (M_0)^(2n)) xx 1/(("mass"))`
Since. g depends upon the 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 third law. Hence, Kepler's third law will not be valid.
As the force is of central nature. .....`[∵ "Force" ∝ 1/r^2]`
Hence, the first two of Kepler's laws will be valid.
For negative n, g = `(GM_0^2 (m_1m_2)^-n)/(r_12^2 (M_0)^(-2n)) xx 1/(("mass"))`
= `(GM_0^(2(1 + n)))/r_12^2 ((m_1m_2)^-n)/(("mass"))`
g = `(GM_0^2)/r_12^2 (M_0^2/(m_1m_2)) xx 1/"mass"`
As M0 > m1 or m2
g > 0, hence in this case situation will reverse i.e., an object lighter than water will sink in water.
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