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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

  • Option 1)

    2R

  • Option 2)

    \frac{R}{\sqrt{2}}

  • Option 3)

    R/2

  • Option 4)

    \sqrt{2}R

 

Answers (1)

best_answer

As we discussed in

Acceleration due to gravity (g) -

Force extended by earth on a body is gravity.

Formula:    g=\frac{GM}{R^{2}},

g=\frac{4}{3}\pi \rho \, GR

g\rightarrow gravity

\rho \rightarrow density of earth

R \rightarrow Radius of earth

 

- wherein

It's average value is 9.8\: m/s^{2}\; \; or \; \; 981cm/sec^{2}\; or\; 32feet/s^{2} on the surface of earth

 

 The accelerator due to gravity at a height h from thegrow of given as \frac{g}{9}

\frac{GM}{r^{2}}=\frac{GM}{R^{2}} \frac{1}{9}

r=3R

The height above the ground is 2R.


Option 1)

2R

Correct option

Option 2)

\frac{R}{\sqrt{2}}

Incorrect option

Option 3)

R/2

Incorrect option

Option 4)

\sqrt{2}R

Incorrect option

Posted by

divya.saini

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