Question

Two particles of mass 1 kg and 2 kg are placed at a seperation of 100 cm then the gravitational force
between them is
[ j
a) 5.3 × 10-N b) 13.34 x 10-11 N c) 2.65 x 10-N d) 7.95 x 10-10 N​

Answer 1

Answer:

The force of gravitastion exerted by one particle on another is

F1=Gm1m2r2

=96.67×10−11N−m2kg2(1.0kg)×(2.0kg))(0.5m)2

=5.3×10−10N

The acceleration of 1.0 kg particle is

a=F/m_1=(5.3xx10^-10N)/(1.0kg)=5.3xx10^-10N ms^2-Thisae≤rationis→wardsthe2.0kgpartic≤.Theae≤rationofthe2.0kgpartic≤isa_2=F/m_2=(5.3xx10^-10N)/(2.0kg)=2.65xx10^-10ms^-2`

This acceleration is towards the 1.0 kg particle.

Answer 2

b) 13.34×10^-11 N

Explanation:

m₁ = 1kg

m₂ = 2kg

d = 100 cm

= 1m

F = ?

Now, F = Gm₁m₂/d²

= 6.67×10^-11 ×1×2 / 1²

= 13.34 ×10^-11 N