Get Answers to all your Questions

header-bg qa

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)


  • Option 2)


  • Option 3)


  • Option 4)



Answers (1)


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}


The height above the ground is 2R.

Option 1)


Correct option

Option 2)


Incorrect option

Option 3)


Incorrect option

Option 4)


Incorrect option

Posted by


View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE