Question

Spheres A and B of uniform density have
masses 1 kg and 100 kg respectively. Their centres are separated by 100 m. (i) Find the gravitational force between them.
(ii) Find the gravitational force on A due to the earth.
(iii) Suppose A and B are initially at rest and A can move freely towards B. What will be the velocity of A one second after it starts moving
towards B?
How will this velocity change with
time?
How much time will A take to move
towards B by 1 cm?
(iv) If A begins to fall,
starting from rest, due to the earth's downward pull, what will be its velocity after one second?
How much time will it take to fall through 1 cm?
[M(earth) = 6 x 1024 kg, earth) = 6400 km]​​

Answer 1

Answer:

${\large{\mathfrak{\red{\clubs{\underline{Given :}}}}}}$

Given:-

• Here first we should write the given data
• m1=1kg,m2=100kg,r=100m,M=6×10^24kg, R= 6400 km=6400×10^2m

• And write everything which is given in question.

To find:-

• F1=?
• F2=?

Explanation :-

1. Lets take first case

• F1= GMm/r^2
• Now applying the values,

• I will write the answer as direct as u can u do calculations

• F1=6.67×10^11N-m^2/kg^2×1kg ×100kg
• F1= 6.67×10^13N.

Case ii)

• F2=Gm1M/R^2
• F2=6.68×10^11N m^2/kg^2×1kg×6×10^24kg

• F2= 40.02×10^12/6.4×6.4×10^12

• N=9.77N.

This is far greater than F1.

Hope it helps u mate .

Thank you .