Question

if simran's weight is 600N on the earth .how far she should be from centre of the earth so that her weight will be 300N?(radius of earth is 64*10^5m)

Answer 1

Given:

• Weight on the Earth surface $\sf{W_{E}}$ is 600N
• Radius (R = 6.4 × ${10}^{6}$ ) = 6400km

To Find:

• Height from centre of the Earth so that her weight will be 300N

Solution:

Let's acceleration above the surface of the Earth be g and g' on the surface of the Earth.

As given weight if simran is 600N

We have to use the formula of weight.

Formula:

$\: \: \sf \: weight = mass \times accceleration \: of \: gravity$

Denote weight is W mass is M acceleration of gravity is g on the Earth surface

$\: \: \sf \: w = m \times g$

600=mg _____(i)

Weight will reduce on the surface of the Earth and if reduce in gravity then also reduce in other quantity.

Weight above the surface:

Since we know that mass will remain same and we know mass does not change it remains constant.

300=mg_____(ii)

$\dfrac{g}{g'}$=$\dfrac{600m}{300m}$

$\dfrac{g}{g'}$=2_______(iii)

Now,we will use the formula for acceleration of gravity

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

$\: \: \sf \: 2 = \frac{ \cancel g \cancel m}{ {r}^{2} } \times \frac{ {(r + h)}^{2} }{ \cancel g \cancel m} \\ \\ \: \sf \: taking \: square \: root \\ \\ \: \: \sf \: \sqrt{2} = \sqrt{ \frac{ {(r + h)}^{2} }{ {r}^{2} } } \\ \\ \: \: \sf \: \sqrt{2} = \frac{r + h}{2} \\ \\ \: \: \sf \: \sqrt{2} r= (r + h) \\ \\ \: \: \sf \: h \frac{r}{1 + \sqrt{2} } \\ \\ \: \: \sf \: h = \frac{6400}{1 + \sqrt{2} } \\ \\ \: \: \sf \: h = \frac{6400}{2.41} \\ \\ \: \: \sf \: h = 2651$

Hence, Height from centre of the Earth so that her weight will 300N is 2651m.

Answer 2

Answer:

here is answer dear friend