Question

two bodies each of mass 1 kg and 2 kg respectively are placed at a separation of 1 m. Find the acceleration of the bodies assuming that only gravitational forces act on them

Answer 1

Answer:

this is correct only at t = 0

Explanation:

if the particles start moving and as they are nearby their distance of separation decreases and thus force is no more constant and hence acceleration will also change

Answer 2

Answer:

Acceleration of 1 kg mass is $1.334 \times 10^{-11}$and 2kg mass is $0.667 \times 10^{-11}$

Step by step Explanation:

Given explanation,

Two bodies each of mass 1 kg and 2 kg respectively.

Two bodies placed at a separation of 1 m.

We have to find out The acceleration of the bodies.

Let,

m1 and m2 are the mass of two bodies .

m1 = 1 kg

m2 = 2 kg

r = 1 m

As we know that,

$F = G\frac{m1\times m2}{r^{2} }$

Where,

G is a gravitational force acting on the mass = $6.67 \times 10^{-11}$

$F = 6.67 \times 10^{-11}\times \frac{1\times 2}{1^{2} }$

Square means multiplying the given value with itself twice.

$1^{2}=1 \times 1 = 1$

$F = 6.67 \times 10^{-11}\times \frac{1\times 2}{1}$

The value of 1 at the denominator is get neglected

$F = 6.67 \times 10^{-11}\times 1\times 2$

$F = 6.67 \times 10^{-11}\times 2$

$F = 1.334 \times 10^{-11}$

As per the Newton's second law

F = m a

Let, a1 be the acceleration of body having mass 1kg and  a2 be the acceleration of body having mass 2kg.

For a1 be the acceleration of body having mass 1kg.

$F = m1 \times a1$

$a1 = \frac{F}{m1}$

$a1 = \frac{1.334 \times 10^{-11}}{1}$

$a1 = 1.334 \times 10^{-11}$

For a2 be the acceleration of body having mass 2kg.

$F = m2 \times a2$

$a2 = \frac{F}{m2}$

$a2 = \frac{1.334 \times 10^{-11}}{2}$

By dividing by 2  we get

$a2 = 0.667 \times 10^{-11}$

Hence,

Acceleration of 1 kg mass is $1.334 \times 10^{-11}$and 2kg mass is $0.667 \times 10^{-11}$ .