Question

At what height above the surface of earth does the acceleration due to gravity become 16% of the value of the surface of the eart

Answer 1

Answer:

To find:

Height at which acceleration due to gravity becomes 16% as compared to Earth surfaces.

Calculation:

Let height be h , radius of Earth be r , gravity at Earth surface be g , gravity at height be g2

$\boxed{ \huge{ \red{ \sf{ \therefore \: g2 = (16\%)g}}}}$

Now, we can say that :

$\star \: g2 = \dfrac{g}{ \bigg \{{1 + (\dfrac{h}{r} ) \bigg \}}^{2} }$

$\implies\: \dfrac{16}{100}g = \dfrac{g}{ \bigg \{{1 + (\dfrac{h}{r} ) \bigg \}}^{2} }$

$\implies\: \dfrac{16}{100} = \dfrac{1}{ \bigg \{{1 + (\dfrac{h}{r} ) \bigg \}}^{2} }$

Taking Square root on both sides:

$\implies\: \dfrac{4}{10} = \dfrac{1}{ \bigg \{{1 + (\dfrac{h}{r} ) \bigg \}} }$

$\implies \: 1 + \dfrac{h}{r} = \dfrac{10}{4}$

$\implies \: \dfrac{h}{r} = \dfrac{10}{4} - 1 = \dfrac{6}{4} = \dfrac{3}{2}$

$\implies \: h = \dfrac{3r}{2}$

So final answer is:

$\boxed{ \huge{ \orange{ \sf{ \: h = \dfrac{3r}{2}}}}}$

Answer 2

$\huge{\underline{\underline{\red{\mathfrak{AnSwEr :}}}}}$

Given :

• At a height, acceleration due to gravity becomes 16% of its original value

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To Find :

• Height at which the acceleration due to gravity becomes 16%

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Solution :

So, A.T.Q

$\large{\boxed{\boxed{\sf{g_2 \: = \: 16 \% g_1}}}}$

Also, in the derivation of Variation of gravitational force due to height above Earth surface. We have

$\large{\boxed{\boxed{\sf{g_2 \: = \: \dfrac{g_1}{\bigg(1 \: + \: \dfrac{h}{r} \bigg) ^2}}}}} \\ \\ \implies {\sf{\dfrac{16}{100}g_1 \: = \:\dfrac{g_1}{\bigg(1 \: + \: \dfrac{h}{r} \bigg) ^2}}} \\ \\ \implies {\sf{\dfrac{16}{100} \: = \: \dfrac{1}{\bigg(1 \: + \: \dfrac{h}{r} \bigg) ^2}}} \\ \\ \small {\gray {\dag {\rm{\underline{\: \: \: \: \: \: \: Square \: root \: both \: sides \: \: \: \: \: \: \: }}}}} \gray \dag \\ \\ \implies {\sf{\dfrac{4}{10} \: = \: \dfrac{1}{1 \: + \: \dfrac{h}{r}} }} \\ \\ \implies {\sf{1 \: + \: \dfrac{h}{r} \: = \: \dfrac{10}{4}}} \\ \\ \implies {\sf{\dfrac{h}{r} \: = \: \dfrac{10}{4} \: - \: 1}} \\ \\ \implies {\sf{h \: = \: \dfrac{3r}{2}}}$

Value of height is 3r/2