Question

At what height 'h' from surface of earth the acceleration due to gravity decreases 50 %

Answer 1

Answer:

At what height 'h' from surface of earth the acceleration due to gravity decreases 50 %

Answer 2

The new value of gravity at a height h is given as g′=

$\frac{g \: {r}^{2} }{(r + h) {}^{2} }$

As visible from the formula that incresing h results in decresing gravity.

So just put g′=g/2 i.e. 50% less than g where g is gravity at surface. so we get g/2=

=

$g \div 2 = \frac{g \: {r}^{2} }{(r + h) {}^{2} } or \: \frac{1}{2} \frac{ \: {r}^{2} }{(r + h) {}^{2} } \\ or \frac{2}{1} = \frac{(r + h) {}^{2} }{ {r}^{2} } or \sqrt[2]{2} = \frac{r + h}{r} \\ = 1 + \frac{h}{r}$

=1+h/R putting values we get

$\frac{h}{r}$

=1.414−1=0.414

so h=0.414R=2649.6Km