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(Radius of earth = \[6.4\times {{10}^{6}}m\])

(a) \[1.6\times {{10}^{6}}m\]

(b) \[6.4\times {{10}^{6}}m\]

(c) \[9.0\times {{10}^{6}}m\]

(d) \[2.6\times {{10}^{6}}m\]

Answer
Verified

Hint: The relation between value of acceleration, radius of earth and height can be given by the formula, ${{g}_{h}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}}$, so using this relation we will find the value of h by equating the values of gravitational acceleration.

Formula used: ${{g}_{h}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}}$

Complete step by step answer:

In question we are given that the value of acceleration due to gravity at Earth’s surface is $9.8m{{s}^{-2}}$. Now, we are asked to find the altitude above its surface at which the acceleration due to gravity decreases to $4.9m{{s}^{-2}}$. The relation between gravitational acceleration, height h and radius R can be given by the formula,

${{g}_{h}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}}$ ……………….(ii)

Where, G is the gravitational constant of earth, m is mass of object, R is radius of earth and h is height of earth.

Now, the gravitational acceleration at surface of earth can be given as,

${{g}_{h}}=\dfrac{GM}{{{\left( R \right)}^{2}}}$ ……………….(ii)

As, height is zero on the surface of the earth.

Now, we are given that value of gravitational acceleration on surface of earth is $9.8m{{s}^{-2}}$ and it is $4.9m{{s}^{-2}}$ at height h, so on substituting these values in expression (ii) and (i) we will get,

\[4.9=\dfrac{GM}{{{\left( R+h \right)}^{2}}}\Rightarrow GM=4.9{{\left( R+h \right)}^{2}}\]

$9.8=\dfrac{GM}{{{\left( R \right)}^{2}}}\Rightarrow GM=9.8{{\left( R \right)}^{2}}$

On equating the equations, we will get,

\[4.9{{\left( R+h \right)}^{2}}=9.8{{\left( R \right)}^{2}}\]

\[\dfrac{{{\left( R+h \right)}^{2}}}{{{\left( R \right)}^{2}}}=\dfrac{9.8}{4.9}\Rightarrow \dfrac{{{\left( R+h \right)}^{2}}}{{{\left( R \right)}^{2}}}=2\]

On taking square roots on both the sides we will get,

\[\dfrac{\left( R+h \right)}{\left( R \right)}=\sqrt{2}\Rightarrow R+h=\sqrt{2}R\]

\[h=\sqrt{2}R-R\Rightarrow h=\left( \sqrt{2}-1 \right)R\]

\[\Rightarrow h=\left( 1.41-1 \right)6.4\times {{10}^{6}}=\left( 0.41 \right)6.4\times {{10}^{6}}\]

\[\Rightarrow h=2.6\times {{10}^{6}}m\]

Thus, the value of altitude will be \[2.6\times {{10}^{6}}m\].

Hence, option (d) is the correct answer.

Note: This problem can also be solved by using the formula of gravitational acceleration at height h which can be given as, ${{g}_{h}}={{g}_{o}}\dfrac{R}{{{\left( R+h \right)}^{2}}}$, where ${{g}_{h}}={{g}_{o}}=9.8m{{s}^{-2}}$ for value for gravitation at surface of earth and ${{g}_{h}}=4.9m{{s}^{-2}}$, for acceleration at height h. And again, equating the expressions we will get the same answer. So, we can say that this is an alternative method to solve this type of problem.

Formula used: ${{g}_{h}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}}$

Complete step by step answer:

In question we are given that the value of acceleration due to gravity at Earth’s surface is $9.8m{{s}^{-2}}$. Now, we are asked to find the altitude above its surface at which the acceleration due to gravity decreases to $4.9m{{s}^{-2}}$. The relation between gravitational acceleration, height h and radius R can be given by the formula,

${{g}_{h}}=\dfrac{GM}{{{\left( R+h \right)}^{2}}}$ ……………….(ii)

Where, G is the gravitational constant of earth, m is mass of object, R is radius of earth and h is height of earth.

Now, the gravitational acceleration at surface of earth can be given as,

${{g}_{h}}=\dfrac{GM}{{{\left( R \right)}^{2}}}$ ……………….(ii)

As, height is zero on the surface of the earth.

Now, we are given that value of gravitational acceleration on surface of earth is $9.8m{{s}^{-2}}$ and it is $4.9m{{s}^{-2}}$ at height h, so on substituting these values in expression (ii) and (i) we will get,

\[4.9=\dfrac{GM}{{{\left( R+h \right)}^{2}}}\Rightarrow GM=4.9{{\left( R+h \right)}^{2}}\]

$9.8=\dfrac{GM}{{{\left( R \right)}^{2}}}\Rightarrow GM=9.8{{\left( R \right)}^{2}}$

On equating the equations, we will get,

\[4.9{{\left( R+h \right)}^{2}}=9.8{{\left( R \right)}^{2}}\]

\[\dfrac{{{\left( R+h \right)}^{2}}}{{{\left( R \right)}^{2}}}=\dfrac{9.8}{4.9}\Rightarrow \dfrac{{{\left( R+h \right)}^{2}}}{{{\left( R \right)}^{2}}}=2\]

On taking square roots on both the sides we will get,

\[\dfrac{\left( R+h \right)}{\left( R \right)}=\sqrt{2}\Rightarrow R+h=\sqrt{2}R\]

\[h=\sqrt{2}R-R\Rightarrow h=\left( \sqrt{2}-1 \right)R\]

\[\Rightarrow h=\left( 1.41-1 \right)6.4\times {{10}^{6}}=\left( 0.41 \right)6.4\times {{10}^{6}}\]

\[\Rightarrow h=2.6\times {{10}^{6}}m\]

Thus, the value of altitude will be \[2.6\times {{10}^{6}}m\].

Hence, option (d) is the correct answer.

Note: This problem can also be solved by using the formula of gravitational acceleration at height h which can be given as, ${{g}_{h}}={{g}_{o}}\dfrac{R}{{{\left( R+h \right)}^{2}}}$, where ${{g}_{h}}={{g}_{o}}=9.8m{{s}^{-2}}$ for value for gravitation at surface of earth and ${{g}_{h}}=4.9m{{s}^{-2}}$, for acceleration at height h. And again, equating the expressions we will get the same answer. So, we can say that this is an alternative method to solve this type of problem.

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