A sphere of mass 40 kg is attracted by another sphere of mass 15 kg with a force of 1-10m g wt- Find the constant of gravitation if the centres of the spheres are 0-2 mapart-

Step 1: GivenMass of 1st sphere, M_1 = 40 kg0ex0exMass of 2nd sphere, M_2 = 15 kg0ex0exGravitation force, F = 1/10m g wt0ex0exDistance between centres of both s ...

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Standard X

Physics

Gravitational Force

A sphere of m...

Question

A sphere of mass $40 k g$ is attracted by another sphere of mass $15 k g$ with a force of $\frac{1}{10} m g w t$ . Find the constant of gravitation if the centres of the spheres are $0.2 m$ apart.

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Solution

Step 1: Given
$M a s s o f 1 s t s p h e r e, M_{1} = 40 k g M a s s o f 2 n d s p h e r e, M_{2} = 15 k g G r a v i t a t i o n f o r c e, F = \frac{1}{10} m g w t D i s \tan c e b e t w e e n c e n t r e s o f b o t h s p h e r e s, R = 0.2 m U n i v e r s a l g r a v i t a t i o n a l c o n s \tan t, G = ?$
Step 2: Formula used
Converting the value of force in SI unit (Newton),
$F = \frac{1}{10} m i l l i g r a m w t = \frac{1}{10} \times 10^{- 3} g w t = \frac{1}{10} \times 10^{- 3} \times 10^{- 3} k g w t = 10^{- 7} \times 9.8 N = 9.8 \times 10^{- 7} N$
Step 3: Calculations
Rearranging the Universal law of gravitation to find the value of G,
$F = G \times \frac{M_{1} \times M_{2}}{R^{2}} G = \frac{F \times R^{2}}{M_{1} \times M_{2}}$
Substituting the values in the expression,
$G = \frac{F \times R^{2}}{M_{1} \times M_{2}} = \frac{9.8 \times 10^{- 7} N \times {(0.2 m)}^{2}}{40 k g \times 15 k g} = 6.533 \times 10^{- 11} N m^{2} k g^{- 2}$
Hence, the value of G is $6.533 \times 10^{- 11} N m^{2} k g^{- 2}$ .

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A sphere of mass 40kg is attracted by another sphere of mass 15kg with a force of 110mgwt. Find the constant of gravitation if the centres of the spheres are 0.2mapart.

A sphere of mass $40 k g$ is attracted by another sphere of mass $15 k g$ with a force of $\frac{1}{10} m g w t$ . Find the constant of gravitation if the centres of the spheres are $0.2 m$ apart.