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The value of acceleration due to gravity at a point P inside the earth and at another point Q outside the earth is g/2 (g being acceleration due to gravity at the surface of earth). Maximum possible distance in terms of radius of earth R between P and Q is?

The value of acceleration due to gravity at a point P inside the earth and at another point Q outside the earth is g/2 (g being acceleration due to gravity at the surface of earth). Maximum possible distance in terms of radius of earth R between P and Q is?

Question Image
Grade:11

3 Answers

Vikas TU
14149 Points
4 years ago
For (for above the surface),
g/2 = GM/(R+h)^2
For below the surface,
g/2 = GM/(R-d)^2

Equating both we get,
R^2 = h^2 + 2Rh
and
R^2 = d^2 – 2Rd
solve quadratic for both h and d terms as:
h = (root2 – 1)R
and
d = (root2 + 1)R
hence answer is:
h + d = (root2 + 1)R + (root2 - 1)R => 2root(2)R meter.
Hasib
15 Points
2 years ago
 (for above the surface),For
g/2 = GM/(R+h)^2
For below the surface,
g/2 = GM/(R-d)^2

Equating both we get,
R^2 = h^2 + 2Rh
and
R^2 = d^2 – 2Rd
solve quadratic for both h and d terms as:
h = (root2 – 1)R
and
d = 3/2R
hence answer is:
h+d= R/2(2*Root2+1)
Rishi Sharma
askIITians Faculty 646 Points
11 months ago
Dear Student,
Please find below the solution to your problem.

For(for above the surface),
g/2 = GM/(R+h)^2
For below the surface,
g/2 = GM/(R-d)^2
Equating both we get,
R^2 = h^2 + 2Rh
and
R^2 = d^2 – 2Rd
solve quadratic for both h and d terms as:
h = (root2 – 1)R
and
d = (root2 + 1)R
hence answer is:
h + d = (root2 + 1)R + (root2 - 1)R
=> 2root(2)R meter.

Thanks and Regards

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