Question

NEWTON'S UNIVERSAL
LAW OF GRAVITATION :
Que 1 : The gravitational force between two bodies
is 6.67x10-?N when the distance between
their centres is 10 m. If the mass of first
body is 800 kg, then the mass of second
body is
1) 1000 kg
2) 1250 kg

Que 2: A 3 kg mass and a 4 kg mass are placed on x
and y axes at a distance of 1 metre from the
origin and a 1 kg mass is placed at the origin.
Then the resultant gravitational force on 1 kg
mass is
1) 7G 2) G 3) 5G

Que 3 : Two particles of equal mass go around in a
circle of radius 'r'under the action of their
mutual gravitational attraction. If the mass /
of each particle is m, the speed of each
particle is?
1) root Gm/r 2)root Gm/4r

p.s I need explanations :)​

Answer 1

1)Given, F = 6.67 X 10-7 N

r = 10 m

m1 = 800 kg

m2 = ?

Using formula :

F = G m1 m2r2where, G = 6.67 × 10−11 N m2 kg−2

Put corresponding values in the above formula and calculate for m2.

2)Assume P is the point where we have a mass of 1 kg. and net force P is zero.

AB = 2m

Let AP = x

2 BP = AB – AP

= (2 – x) m

Since net force at P is zero.

So, FPA = FPB

⇒ 2 – x = 2x

⇒ 3x = 2

3]Assume that three particles are at points A, B and C on the circumference of a circle.

BC = CD = 2√a

The force on the particle at C due to gravitational attraction of the particle at B is F⃗ CB=GM22R2jˆ.

The force on the particle at C due to gravitational attraction of the particle at D is F⃗ CD=−GM22R2iˆ.

Now, force on the particle at C due to gravitational attraction of the particle at A is given by

F⃗ CA=−GM24R2cos 45iˆ+GM24R2sin 45 jˆ∴F⃗ C=F⃗ CA+F⃗ CB+F⃗ CD =−GM24R2(2+12√)iˆ+GM24R2(2+12√) jˆ

So, the resultant gravitational force on C is FC=Gm24R222√+1−−−−−−−√.

Let v be the velocity with which the particle is moving.

Centripetal force on the particle is given by

F=mv2R ⇒v=GMR(22√+14)−−−−−−−−−−−√