Asked by chris neives
Two 0.61 kg basketballs, each with a radius of 13 cm, are just touching. How much energy is required to change the separation between the centers of the basketballs to the following (a)1 meter
(b) 10 meters
(b) 10 meters
Answers
Answered by
Elena
the gravitational constant G =6.67•10⁻¹¹ N•m²/kg²,
PE₁=G•m₁•m₂/R₁
PE₂=G•m₁•m₂/R₂
E=ΔPE= PE₁-PE₂=
=G•m₁•m₂/R₁-G•m₁•m₂/R₂=
=G•m₁•m₂(1/ R₁- 1/R₂)= …
R₁=0.26 m
R₂= 1 m, 10 m
PE₁=G•m₁•m₂/R₁
PE₂=G•m₁•m₂/R₂
E=ΔPE= PE₁-PE₂=
=G•m₁•m₂/R₁-G•m₁•m₂/R₂=
=G•m₁•m₂(1/ R₁- 1/R₂)= …
R₁=0.26 m
R₂= 1 m, 10 m
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