Science, asked by maheshbhatt179, 4 months ago

if the weight of an object is 40 n, what will be its weight when taken at a distance 2R from the center of the earth​

Answers

Answered by Anonymous
2

GIVEN :-

  • Weight of the object is 40N at the surface with radius R.

 \\

TO FIND :-

  • Weight of the object when it is taken at distance 2R from the centre of earth.

 \\

TO KNOW :-

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

G is Gravitational constant.

 \\   \bigstar\boxed{ \sf \:F =  \dfrac{GM(1)M(2)}{ {R}^{2} }  } \\

Here ,

  • F → Weight / Force.
  • G → Gravitational constant.
  • M(1) → Mass of object.
  • M(2) → Mass of earth.
  • R → Radius of earth.

 \\

SOLUTION :-

We know ,

 \\  \sf \: F =  \dfrac{GM(1)M(2)}{ {R}^{2} }  \\

Here , G , M(1) and M(2) are constant.

Therefore ,

 \\  \sf \: F \propto \dfrac{1}{ {R}^{2} }  \\  \\  \sf \: F {R}^{2}  = constant \\

Hence ,

 \\   \boxed{\boxed{ \sf \:  F_{1} { R_{1} }^{2}  =  F_{2} { R_{2} }^{2} }} \\

We have ,

  • F{1} = 40N
  • R{1} = R
  • F{2} = ?
  • R{2} = 2R

 \\

Putting values we get,

 \\  \implies \sf \: 40( {R)}^{2}  =  F_{2} ( {2R)}^{2}  \\  \\ \implies  \sf \: 40 {R}^{2}  = F_{2} (4 {R)}^{2}  \\  \\  \implies \sf \: F_{2}  =   \cancel\dfrac{40 {R}^{2} }{4 {R}^{2} }  \\  \\    \implies \underbrace{ \boxed{\sf \: F_{2}  = 10N} }\\  \\

Hence , weight of object is 10N when it is taken to distance 2R from the core of earth.

Answered by saeeatharv2005
0

Answer:

GIVEN :-

Weight of the object is 40N at the surface with radius R.

\begin{gathered} \\ \end{gathered}

TO FIND :-

Weight of the object when it is taken at distance 2R from the centre of earth.

\begin{gathered} \\ \end{gathered}

TO KNOW :-

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

G is Gravitational constant.

\begin{gathered} \\ \bigstar\boxed{ \sf \:F = \dfrac{GM(1)M(2)}{ {R}^{2} } } \\ \end{gathered}

F=

R

2

GM(1)M(2)

Here ,

F → Weight / Force.

G → Gravitational constant.

M(1) → Mass of object.

M(2) → Mass of earth.

R → Radius of earth.

\begin{gathered} \\ \end{gathered}

SOLUTION :-

We know ,

\begin{gathered} \\ \sf \: F = \dfrac{GM(1)M(2)}{ {R}^{2} } \\ \end{gathered}

F=

R

2

GM(1)M(2)

Here , G , M(1) and M(2) are constant.

Therefore ,

\begin{gathered} \\ \sf \: F \propto \dfrac{1}{ {R}^{2} } \\ \\ \sf \: F {R}^{2} = constant \\ \end{gathered}

F∝

R

2

1

FR

2

=constant

Hence ,

\begin{gathered} \\ \boxed{\boxed{ \sf \: F_{1} { R_{1} }^{2} = F_{2} { R_{2} }^{2} }} \\ \end{gathered}

F

1

R

1

2

=F

2

R

2

2

We have ,

F{1} = 40N

R{1} = R

F{2} = ?

R{2} = 2R

\begin{gathered} \\ \end{gathered}

Putting values we get,

\begin{gathered} \\ \implies \sf \: 40( {R)}^{2} = F_{2} ( {2R)}^{2} \\ \\ \implies \sf \: 40 {R}^{2} = F_{2} (4 {R)}^{2} \\ \\ \implies \sf \: F_{2} = \cancel\dfrac{40 {R}^{2} }{4 {R}^{2} } \\ \\ \implies \underbrace{ \boxed{\sf \: F_{2} = 10N} }\\ \\ \end{gathered}

⟹40(R)

2

=F

2

(2R)

2

⟹40R

2

=F

2

(4R)

2

⟹F

2

=

4R

2

40R

2

F

2

=10N

Hence , weight of object is 10N when it is taken to distance 2R from the core of earth.

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