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
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.
Here ,
- F → Weight / Force.
- G → Gravitational constant.
- M(1) → Mass of object.
- M(2) → Mass of earth.
- R → Radius of earth.
SOLUTION :-
We know ,
Here , G , M(1) and M(2) are constant.
Therefore ,
Hence ,
We have ,
- F{1} = 40N
- R{1} = R
- F{2} = ?
- R{2} = 2R
Putting values we get,
Hence , weight of object is 10N when it is taken to distance 2R from the core of earth.
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.