The value of acceleration due to gravity at

earth's surface is 9.8 ms^{-2}. The altitude

above its surface at which the acceleration

due to gravity decreases to 4.9 ms^{-2}, is

close to : (Radiuas of earth = 6.4 x 10^{^{6}} m)

  • Option 1)

    2.6 X 10^{6} m

  • Option 2)

    6.4 x 10^{6} m

  • Option 3)

    9.0 x 10^{6} m

  • Option 4)

    1.6 x 10^{^{6}} m

 

Answers (1)

 

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

 

 

Variation in 'g' with height -

 

 

g'\rightarrow gravity at height h from surface of earth.

R\rightarrow Radius of earth

h\rightarrow height above surface
 

- wherein

g'\alpha\, \frac{1}{r^{2}}

r=R+h

 

 

g=\frac{GM}{R^{2}}

g'=\frac{g}{2}=\frac{GM}{(R+h)^{2}}

=>\frac{1}{2}=\frac{R^{2}}{(R+h)^{2}}

=> R+h=\sqrt 2 \: \: R

=> R=\frac{h}{\sqrt2-1}

=> h=(\sqrt2-1) R

=> h=0.41\times 6.4\times 10^{6}m

=> h=2.6\times 10^{6}m


Option 1)

2.6 X 10^{6} m

Option 2)

6.4 x 10^{6} m

Option 3)

9.0 x 10^{6} m

Option 4)

1.6 x 10^{^{6}} m

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