Question

here is ur 15 points questions

at what height from the centre of the earth the acceleration due to gravity will be 1/4th of its value as at the earth

...with full explanation...



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Answer 1

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

As we know that,

Variation of g with height :-

$\tt{\rightarrow g_{h}=g(\dfrac{R}{R+h})^2 ......(1)}$

Now,

To Find :-

Height (h) = ?

According to the information provided in question :-

$\tt{\rightarrow g_{h}=\dfrac{1}{4}g......(2)}$

Now,

Substitute value of (2) in (1) we get :-

$\tt{\rightarrow\dfrac{g}{4}=g(\dfrac{R}{R+h})^2}$

$\tt{\rightarrow\dfrac{1}{4}=(\dfrac{R}{R+h})^2}$

Now,

Squaring both sides :-

$\tt{\rightarrow\sqrt{\dfrac{1}{4}}=\sqrt{(\dfrac{R}{R+h})^2}}$

$\tt{\rightarrow\dfrac{1}{2}=\dfrac{R}{R+h}}$

R + h = 2R

h = 2R - R

h = R

Hence,

When height = Radius then g will be $\tt{\dfrac{1}{4}}$ of it's value of surface of Earth.

Answer 2

Answer:

According to the information provided in question :-

\tt{\rightarrow g_{h}=\dfrac{1}{4}g......(2)}→g

h

=

4

1

g......(2)

Now,

Substitute value of (2) in (1) we get :-

\tt{\rightarrow\dfrac{g}{4}=g(\dfrac{R}{R+h})^2}→

4

g

=g(

R+h

R

)

2

\tt{\rightarrow\dfrac{1}{4}=(\dfrac{R}{R+h})^2}→

4

1

=(

R+h

R

)

2

Now,

Squaring both sides :-

\tt{\rightarrow\sqrt{\dfrac{1}{4}}=\sqrt{(\dfrac{R}{R+h})^2}}→

4

1

=

(

R+h

R

)

2

\tt{\rightarrow\dfrac{1}{2}=\dfrac{R}{R+h}}→

2

1

=

R+h

R

R + h = 2R

h = 2R - R

h = R

Hence,

When height = Radius then g will be \tt{\dfrac{1}{4}}

4

1

of it's value of surface of Earth.