Question

A sphere of mass 40kg is attracted by a second spare of mass 60 kg with a force equal

to 4 x 10-5 N of G = 6 x 10-11 Nm2/kg2 calculate the distance B/w the two spheres.​

Answer 1

GIVEN :-

• Mass of Sphere(1) = 40kg

• Mass of Sphere(2) = 60kg

• F = 4 × 10¯⁵ N

• G = 6 × 10¯¹¹

TO FIND :-

• Distance between the two spheres.

TO KNOW :-

Force of attraction between the bodies is directly proportional to the product of their masses and inversely proportional to the square of distance between them.

$\\ \bigstar \boxed{\sf \: F= G\frac{ m_{1} . m_{2}}{ {r}^{2} } }$

Here ,

• F = Force of attraction

• G = Gravitational constant

• m{1} = Mass of object 1

• m{2} = Mass of object 2

• r = Distance between the two objects.

SOLUTION :-

We have ,

• F = 4 × 10¯⁵ N
• G = 6 × 10¯¹¹
• m{1} = 40kg
• m{2} = 60kg
• r = ?

Putting values in the formula ,

$\\ \\ \sf \: 4 \times {10}^{ - 5} = 6 \times {10}^{ - 11} \left( \dfrac{40 \times 60}{ {r}^{2} } \right) \\ \\ \\ \sf \: \dfrac{4 \times {10}^{ - 5} }{6 \times {10}^{ - 11} } = \dfrac{2400}{ {r}^{2} } \\ \\ \\ \sf \: 0.66 \times {10}^{ - 5 - ( - 11)} = \dfrac{2400}{ {r}^{2} } \\ \\ \\ \sf \: 0 .66 \times {10}^{ - 5 + 11} = \dfrac{2400}{ {r}^{2} } \\ \\ \\ \sf \: 0. 66 \times {10}^{6} = \dfrac{2400}{ {r}^{2} } \\ \\ \\ \sf \: {r}^{2} = \dfrac{2400}{0 .66 \times {10}^{6} } \\ \\ \\ \sf \: {r}^{2} = \dfrac{ \cancel{2400}}{ \cancel{66 }\times {10}^{4} } \\ \\ \\ \sf {r}^{2} = 36.36 \times {10}^{ - 4} \\ \\ \\ \sf \boxed{ \sf r = 6.02 \times {10}^{ - 2} m}$

Hence , distance between the two spheres is 6.02×10¯² m or 6.02cm.