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The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is : (a) R / v2 (b) R / 2 (c) v2 R (d) 2 R

The height at which the acceleration due to gravity becomes g/9 (where g = the acceleration due to gravity on the surface of the earth) in terms of R, the radius of the earth, is :
 
(a) R / v2
(b) R / 2
(c) v2 R
(d) 2 R

Grade:6

3 Answers

Deepak Patra
askIITians Faculty 471 Points
9 years ago

(d) We know that g’/g = R2/(R+h)2

? g/9/g = [R/R+h]2 ? h = 2R
Manas Satish Bedmutha
22 Points
9 years ago
By Newton’s law of gravitation, g is inversely proportional to r2,
So, g‘ / g = [R / R+h ] 2
Put g’ = g/9;  h= 2R
Kushagra Madhukar
askIITians Faculty 628 Points
3 years ago
Dear student,
Please find the solution to your problem.
 
We have the relation, gh = g/(1 + h/R)2
Hence, (1 + h/R)2 = g/(g/9) = 9
hence, 1 + h/R = 3
or, h = 2R
 
Thanks and regards,
Kushagra

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