Mahendra and Virat are sitting at a distance of 1 metre from each other their masses are 75 kg and 80 kg respectively. what is the gravitational force between them?
now , In the above example, assuming
that the bench on which Mahendra is sitting
is frictionless, starting with zero velocity,
what will be Mahendra's velocity of motion
towards Virat after 1 s? Will this velocity
change with time and how?
give explanation step-by-step..
Answers
Answer:
The gravitation force is 4.002×10^-9 N
Mahendra's velocity will be 5.336×10^-7 m/s after 1 second.
Yes Mahendra's velocity will increase with time(why? you will find out in explanation..)
Explanation:
According to universal law of gravitation every object attracts all other object in the universe and as heavier and heavier they are the more force will be the force of attraction and as nearer and nearer they get the force between them also increases and Newton provided a formula which is:
(Gm1m2)/r²
so G=6.67×10^-11
m1= 75 kg
m2=80 kg
r= 1 metre
so the gravitational force is
6.67×10^-11×75×80/1²
= 6.67×10^-11×6000
=6×1000×6.67×10^-11
=6×6.67×10^-8
=40.02×10^-8
or 4.002×10^-9N
Velocity that Mahendra acquires is
mass of Mahendra ×acceleration = the gravitation force we got earlier
so 75× (v-u)/t= 4.002×10^-9
since u=0 hence
75v/t=4.002×10^-9
time is given i.e. 1 second
so 75v= 4.002×10^-9
= v=( 4.002×10^-9)/75
=v= 0.05336×10^-9
.:v= 5.336×10^-7 m/s
Now as Mahendra move towards Virat with velocity of 5.336×10^-7 the distance become smaller and smaller and as discussed earlier ,according to universal law of gravitation every object attracts all other object in the universe and the nearer and nearer they get the more and more the force of attraction will be felt by them..