prepare tablee
work sheet
6
values for the following polygon
als
Ople) -22 150p)-2°+3x +266). 2.
Oply) = x 23-5-6
Answers
Answer :
Gravitational force of attraction between the two bodies is 16.675 × 10⁻¹⁴ N.
Explanation :
Given :
Mass of the two spheres, m₁ = m₂ = 500 g or 0.5 kg
Distance of seperation between them, d = 1 m.
Universal gravitation constant, G = 6.67 × 10⁻¹¹ N m² kg⁻²
To find :
Gravitational force of attraction between the two bodies, g = ?
Knowledge required :
Formula for Gravitational force of attraction :
\boxed{\sf{g = G\dfrac{m_{1}m_{2}}{d^{2}}}}
g=G
d
2
m
1
m
2
Where,
g = Acceleration due to gravity
m₁ = Mass of the body
m₂ = Mass of the body
d = Distance of seperation between the two bodies.
Solution :
By using the formula for Gravitational force of attraction and substituting the values in it, we get :
\begin{gathered}:\implies \sf{g = G\dfrac{m_{1}m_{2}}{d^{2}}} \\ \\ :\implies \sf{g = 6.67 \times 10^{-11} \times \dfrac{0.5 × 0.5}{1^{2}}} \\ \\ :\implies \sf{g = 6.67 \times 10^{-11} \times 0.5 × 0.5} \\ \\ :\implies \sf{g = 6.67 \times 10^{-11} \times 0.5^{2}} \\ \\ :\implies \sf{g = 6.67 \times 10^{-11} \times 0.25} \\ \\ :\implies \sf{g = 6.67 \times 10^{-11} \times 2.5 × 10^{-3}} \\ \\ :\implies \sf{g = 16.675 \times 10^{[-11 + (-3)]}} \\ \\ :\implies \sf{g = 16.675 \times 10^{-14}} \\ \\ \boxed{\therefore \sf{g = 16.675 \times 10^{-14}\:N}} \\ \\ \end{gathered}
:⟹g=G
d
2
m
1
m
2
:⟹g=6.67×10
−11
×
1
2
0.5×0.5
:⟹g=6.67×10
−11
×0.5×0.5
:⟹g=6.67×10
−11
×0.5
2
:⟹g=6.67×10
−11
×0.25
:⟹g=6.67×10
−11
×2.5×10
−3
:⟹g=16.675×10
[−11+(−3)]
:⟹g=16.675×10
−14
∴g=16.675×10
−14
N
Therefore,
Gravitational force of attraction between the two bodies, g = 16.675 × 10⁻¹⁴ N.