Question

if your weight is 600N on earth surface...how far should u go from the surface of earth so that your weight will be 300N?(RADIUS =6.4×10^6)??solve this..​

Answer 1

Given :

• Weight on earth, $\sf{W_{E}}$ = 600 N.
• Radius, R = $\sf{6.4\:\times\:10^6}$ = 6400 km.

To Find :

• Height at which the weight will reduce to 300 N.

Solution :

The acceleration produced due to gravity be g and g' on the surface of the earth.

Let the mass be M.

We have the weight of the person on earth's surface given as 600 N.

We have the formula of weight.

Formula :

$\large{\boxed{\sf{\red{Weight\:=\:mass\:\times\:acceleration\:of\:gravity}}}}$

Denote Weight by W, mass by M and acceleration of gravity by g.

Block in the data,

$\sf{W\:=\:m\:\times\:g}$

$\sf{600=\:mg}$ ___(1)

Weight will reduce by half the weight above earth's surface. Reduce in weight will also cause reduction in other quantities.

Weight above earth's surface :

Let the acceleration of gravity above the earth's surface be g'

The mass will remain the same. Since we know, mass doesn't change, it remains constant.

$\sf{300\:=\:mg'}$ ___(2)

From (1) and (2),

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

$\sf{\dfrac{g}{g'}\:=\:2}$ ___(3)

Now, we will use the formula for acceleration of gravity.

$\red{\implies}$ $\sf{\dfrac{g}{g'}\:=\:\frac{\frac{GM}{R^2}}{\frac{GM}{(R+h)^2}}}$

$\red{\implies}$ $\sf{2\:=\:\dfrac{\:\cancel{GM}}{R^2}\:\times\:\dfrac{(R+h)^2}{\:\cancel{GM}}}$

$\red{\implies}$ $\sf{2\:=\:\dfrac{(R+h)^2}{R^2}}$

Taking square roots,

$\red{\implies}$ $\sf{\sqrt{2}\:=\:\sqrt{\dfrac{(R+h)^2}{R^2}}}$

$\red{\implies}$ $\sf{\sqrt{2}\:=\:\dfrac{(R+h)}{R}}$

$\red{\implies}$ $\sf{\sqrt{2}R\:=\:(R+h)}$

$\red{\implies}$ $\sf{h\:=\:\dfrac{R}{(1+\sqrt{2)}}}$

$\red{\implies}$ $\sf{h=\dfrac{6400}{1+\sqrt{2}}}$

$\red{\implies}}$ $\sf{h=\dfrac{6400}{2.41}}$

$\red{\implies}$ $\sf{h=2651}$

$\large{\boxed{\sf{\purple{Height\:at\:which\:the\:weight\:will\:be\:300\:N\:=\:2651\:km}}}}$

Answer 2

Answer:

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