Supposing Newton’s law of gravitation for gravitation forces F1 and F2 between two masses m1 and m2 at positions r1 and r2 read F1=-F2=- r 12 r 12 3 G M o 2 m 1 m 2 M o 2 n where Mo is a constant of dimension of mass, r12 = r1 – r2 and n is a number. In such a case,

(a) The acceleration due to gravity on earth will be different for different objects.

(b) None of the three laws of Kepler will be valid.

(c) Only the third law will become invalid.

(d) For n negative, an object lighter than water will sink in water.

0 1 View | Posted 4 months ago
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  • A

    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    This is a Multiple Choice Questions as classified in NCERT Exemplar

    Answer-a,c,d

    Explanation- F1=-F2=- r 12 r 12 3 G M o 2 m 1 m 2 M o 2 n

    r12=r1-r2

    acceleration due to gravity ,g = F/mass= G M o 2 ( m 1 m 2 ) n r 12 2 ( M o ) 2 n * 1 m a s s

    as g is constant hence constant of proportionality will not be constant in kepler's third law.

    Hence kepler's law will not be valid.

    For negative n, g= G M o 2 m 1 m 2 - n r 12 2 M o - 2 n * 1 m a s s

    = G M o 2 ( 1 + n ) r 12 2 ( m 1 m 2 ) - n m a s s

    Mo>m1 or m2 so objects lighter than water will sink

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