Question

Which of the them is true for two bodies separated by some distance?

a) When the distance between them is halved, Gravitational force

becomes 4 times

b) When one of the mass becomes halved, Gravitational force becomes

halved
c) When the distance between them is increased four times, Gravitational

force becomes 1/16 times

d) None of the above​

Answer 1

Gravitational force is given as follows :

Explanation:

The gravitational force between two masses is given by :

$F=G\dfrac{m_1m_2}{r^2}$

G is Gravitational constant

r is distance between masses

If the distance between them is halved, r' = r/2

(a) New force becomes :

$F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(r/2)^2}\\\\F'=4\times G\dfrac{m_1m_2}{r^2}\\\\F'=4\times F$

So, the gravitational force between masses becomes 4 times of the initial force.

(b) Gravitational force :

$F=G\dfrac{m_1m_2}{r^2}$

If $m'_1=\dfrac{m_1}{2}$

New force,

$F'=G\dfrac{m_1'm_2}{r^2}\\\\F'=G\dfrac{(m_1/2)m_2}{r^2}\\\\F'=\dfrac{1}{2}\times G\dfrac{m_1m_2}{r^2}\\\\F'=\dfrac{1}{2}\times F$

New force becomes half of the initial force.

(c) $F=G\dfrac{m_1m_2}{r^2}$

r' = 4r

New force,

$F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(4r)^2}\\\\F'=G\dfrac{m_1m_2}{16r^2}\\\\F'=\dfrac{1}{16}\times G\dfrac{m_1m_2}{r^2}\\\\F'=\dfrac{1}{16}\times F$

Hence, this is the required solution.

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Answer 2

Answer:

c

Explanation:

the correct answer is c as per the given question.