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.

Answer 1

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.