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Karnataka Board PUCPUC Science Class 11

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

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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.

  1. the acceleration due to gravity on earth will be different for different objects.
  2. none of the three laws of Kepler will be valid.
  3. only the third law will become invalid.
  4. for n negative, an object lighter than water will sink in water.
Short/Brief Note
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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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Chapter 8: Gravitation - Exercises [Page 60]

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NCERT Exemplar Physics [English] Class 11
Chapter 8 Gravitation
Exercises | Q 8.14 | Page 60

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