Question

Question:-

Q. Calculate the force between two spheres of mass 20kg and 30kg are placed 30cm apart ?

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Answer 1

${\rm{\red{\underline{\underline{\huge{Answer}}}}}}$

★ Concept :-

Here the concept Newton's Law of Gravitation has been used. We know that, according to Newton's Law of Gravitation force by which two bodies in universe are attracted is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So we are given with the mass of two bodies and distance between them. So using the formula of Newton's Law of Gravitation, we can find the answer.

Let's Do it:-

__________________________________

Formula Used:-

$\;\boxed{\sf{\pink{F\;=\;\bf{G\:\dfrac{m_{1}\:m_{2}}{d^{2}}}}}}$

★solution:-

Given,

» Mass of first sphere = m₁ = 20 Kg

» Mass of second sphere = m₂ = 30 Kg

» Gravitational Constant = G = 6.67 × 10⁻¹¹ Nm² Kg⁻²

» Distance between two spheres = d = 30 cm = 0.3 m

• Let the Force between the spheres be

By Newton's Law of Gravitation, we know that

$\;\sf{\rightarrow\;\;F\;=\;\bf{G\dfrac{m_{1}\:m_{2}}{d^{2}}}}$

By applying the values, we get

$\;\sf{\Longrightarrow\;\;\green{F\;=\;\bf{G\:\dfrac{20\:\times\:30}{(0.3)^{2}}}}}$

$\;\sf{\Longrightarrow\;\;F\;=\;\bf{G\:\dfrac{20\:\times\:30}{0.3\:\times\:0.3}}}$

$\;\sf{\Longrightarrow\;\;F\;=\;\bf{G\:\dfrac{20\:\times\:100}{0.3}}}$

By applying the value of G, we get

$\;\sf{\Longrightarrow\;\;F\;=\;\bf{\dfrac{6.67\:\times\:10^{-11}\:\times\:20\:\times\:100}{0.3}}}$

$\;\sf{\Longrightarrow\;\;F\;=\;\bf{22.23\:\times\:10^{-11}\:\times\:20\:\times\:100}}$

$\;\sf{\Longrightarrow\;\;F\;=\;\bf{44460\:\times\:10^{-11}}}$

$\;\sf{\Longrightarrow\;\;F\;=\;\bf{44.46\:\times\:10^{-8}}}$

$\;\bf{\Longrightarrow\;\;F\;=\;\bf{\pink{44.5\:\times\:10^{-8}\;\:N}}}$

This Is The Required Answer :-

$\;\underline{\boxed{\tt{Force\;=\;\bf{\purple{44.5\:\times\:10^{-8}\;\:N}}}}}$

★Moretoknow:-

$\;\tt{\leadsto\;\;M\;=\;\dfrac{g\:R^{2}}{G}}$

$\;\tt{\leadsto\;\;g\;=\;\dfrac{G\:M}{R^{2}}}$

$\;\tt{\leadsto\;\;E\;=\;\dfrac{G\:M}{r^{2}}}$

$\;\tt{\leadsto\;\;v_{e}\;=\;\sqrt{\dfrac{G2\:M}{R}}}$

$\;\tt{\leadsto\;\;F\;=\;\dfrac{G\:M\:m_{g}}{R^{2}}}$

Answer 2

Answer:

{\rm{\red{\underline{\underline{\huge{Answer}}}}}}

Answer

★ Concept :-

Here the concept Newton's Law of Gravitation has been used. We know that, according to Newton's Law of Gravitation force by which two bodies in universe are attracted is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. So we are given with the mass of two bodies and distance between them. So using the formula of Newton's Law of Gravitation, we can find the answer.

Let's Do it:-

__________________________________

Formula Used:-

\;\boxed{\sf{\pink{F\;=\;\bf{G\:\dfrac{m_{1}\:m_{2}}{d^{2}}}}}}

F=G

d

2

m

1

m

2

★solution:-

Given,

» Mass of first sphere = m₁ = 20 Kg

» Mass of second sphere = m₂ = 30 Kg

» Gravitational Constant = G = 6.67 × 10⁻¹¹ Nm² Kg⁻²

» Distance between two spheres = d = 30 cm = 0.3 m

• Let the Force between the spheres be

By Newton's Law of Gravitation, we know that

\;\sf{\rightarrow\;\;F\;=\;\bf{G\dfrac{m_{1}\:m_{2}}{d^{2}}}}→F=G

d

2

m

1

m

2

By applying the values, we get

\;\sf{\Longrightarrow\;\;\green{F\;=\;\bf{G\:\dfrac{20\:\times\:30}{(0.3)^{2}}}}}⟹F=G

(0.3)

2

20×30

\;\sf{\Longrightarrow\;\;F\;=\;\bf{G\:\dfrac{20\:\times\:30}{0.3\:\times\:0.3}}}⟹F=G

0.3×0.3

20×30

\;\sf{\Longrightarrow\;\;F\;=\;\bf{G\:\dfrac{20\:\times\:100}{0.3}}}⟹F=G

0.3

20×100

By applying the value of G, we get

\;\sf{\Longrightarrow\;\;F\;=\;\bf{\dfrac{6.67\:\times\:10^{-11}\:\times\:20\:\times\:100}{0.3}}}⟹F=

0.3

6.67×10

−11

×20×100

\;\sf{\Longrightarrow\;\;F\;=\;\bf{22.23\:\times\:10^{-11}\:\times\:20\:\times\:100}}⟹F=22.23×10

−11

×20×100

\;\sf{\Longrightarrow\;\;F\;=\;\bf{44460\:\times\:10^{-11}}}⟹F=44460×10

−11

\;\sf{\Longrightarrow\;\;F\;=\;\bf{44.46\:\times\:10^{-8}}}⟹F=44.46×10

−8

\;\bf{\Longrightarrow\;\;F\;=\;\bf{\pink{44.5\:\times\:10^{-8}\;\:N}}}⟹F=44.5×10

−8

N

This Is The Required Answer :-

\;\underline{\boxed{\tt{Force\;=\;\bf{\purple{44.5\:\times\:10^{-8}\;\:N}}}}}

Force=44.5×10

−8

N

★Moretoknow:-

\;\tt{\leadsto\;\;M\;=\;\dfrac{g\:R^{2}}{G}}⇝M=

G

gR

2

\;\tt{\leadsto\;\;g\;=\;\dfrac{G\:M}{R^{2}}}⇝g=

R

2

GM

\;\tt{\leadsto\;\;E\;=\;\dfrac{G\:M}{r^{2}}}⇝E=

r

2

GM

\;\tt{\leadsto\;\;v_{e}\;=\;\sqrt{\dfrac{G2\:M}{R}}}⇝v

e

=

R

G2M

\;\tt{\leadsto\;\;F\;=\;\dfrac{G\:M\:m_{g}}{R^{2}}}⇝F=

R

2

GMm

g