Two identical solid copper spheres of radius R are placed in a contact with each other then the gravitational attraction between them is proportional to r^x. Then r^x is?
Answers
Answered by
74
gravitational attraction between two objects =
F = G M1 M2 / d²
where d = distance between the centers of mass of the two objects
= 2 R
F = G M² / (2R)² = (G M²/4) R⁻²
So the force α (2 R)⁻² or R⁻²
x = - 2
F = G M1 M2 / d²
where d = distance between the centers of mass of the two objects
= 2 R
F = G M² / (2R)² = (G M²/4) R⁻²
So the force α (2 R)⁻² or R⁻²
x = - 2
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But the answer is 4
Answered by
77
F =GM1M2/R^2
MASS = DESITY*VOLUME
SO M1= D*(4/3 PI R^3)
M2=D*(4/3 PI R^3)
GIVEN RADIUS OF BOTH SPHERES "R" THEY ARE ATTACHED, SO TOTAL RADIUS IS "2R"
F = {G(D*4/3 PI R^3)(D*4/3 PI R^3)}/(2R)^2
F = CONSTANT PART (R^6/4R^2)
F = (CONSTANT PART/4) (R^4)
F = CONSTANT (R^4)
SO THAT ATTRACTION DEPENDS ON R^4
HOPE YOU UNDERSTAND STAND THANK YOU
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