Question

A sphere of mass 40kg is attracted by another sphere of mass 15 kg when their centres are 0.2m apart with force 9.8×10-7 calucalte the value G

Answer 1

Answer:

Distance (G)=20cm=0.2m

Mass (M

1

)=40kg Mass (m

2

)=15kg

Force (F)=0.1 milligram wt

=0.1×10

−3

×980 dyne

=98×10

−3

dyne

=98×10

−8

N

=9.8×10

−9

N

F=

R

2

GM

1

M

2

G=

M

1

M

2

Fk

2

=

40×15

9.8×10

−7

×0.2×0.2

=

4×15

98×2×2

×

10×10×10×10

10

−7

=6.5×10

−11

m

2

/kss

2

solution

Answer 2

Given :

Mass of one sphere (M₁) = 40 Kg

Mass of another sphere (M₂) = 15 Kg

Distance between their centres = 0.2 m

Force = 9.8 × 10⁻⁷ N

To Find :

Value of G

Solution :

We know, the Gravitational Force acting between two masses is equal to $\frac{GM_1M_2}{r^2}$

As per question,

  Gravitational Force (F) =  9.8 × 10⁻⁷

⇒  $\frac{GM_1M_2}{r^2}$                         =  9.8 × 10⁻⁷

⇒  $\frac{G * 40 * 15}{(0.2)^2}$                        =   9.8 × 10⁻⁷

⇒     G                             =   $\frac{9.8 * 10^-^7 * 0.2 * 0.2} {40 * 15}$

⇒     G                             =  $\frac{0.392 * 10^-^7}{600}$

⇒     G                             =  0.0653 × 10⁻⁹ Nm²/Kg²

∴      G                             =  6.53 × 10⁻¹¹ Nm²/Kg²

Therefore, the value of G is found to be 6.53 × 10⁻¹¹ Nm²/Kg²