
How far away from the surface of the earth does the acceleration due to gravity become\[4\;\% \] of its value on the surface of earth? \[\left[ {{R_e} = 6400\;{\rm{km}}} \right]\]
Hint: \[{g_1} = g{\left[ {\dfrac{R}{{R + h}}} \right]^2}\]
Answer
509.4k+ views
Hint: The above problem can be resolved by applying the mathematical formula for the variation of gravitational acceleration when an object is taken to a certain height from the surface of the earth. Moreover, as it is given that the value of gravitational acceleration will become 4% of the value on the surface. Then the further solution is achieved by substituting the values of variables in the mathematical formula.
Complete step by step answer:
The Acceleration due to gravity at height h is given as,
\[{g_1} = g{\left( {\dfrac{{{R_e}}}{{{R_e} + h}}} \right)^2}..............................\left( 1 \right)\]
Here, \[g\] is the gravitational acceleration on the surface of earth and \[{R_e}\] is the radius of earth.
As the gravitational acceleration is 4 % of the value on the surface of the earth. Then the value is given as,
\[{g_1} = \dfrac{{4g}}{{100}}\]
Substituting the value in equation 1 as,
\[\begin{array}{l}
{g_1} = g{\left( {\dfrac{{{R_e}}}{{{R_e} + h}}} \right)^2}\\
\left( {\dfrac{{4g}}{{100}}} \right) = g{\left( {\dfrac{{{R_e}}}{{{R_e} + h}}} \right)^2}\\
0.04 = {\left( {\dfrac{{{R_e}}}{{{R_e} + h}}} \right)^2}\\
0.2 = \dfrac{{{R_e}}}{{{R_e} + h}}
\end{array}\]
Further solving the above equation as,
\[\begin{array}{l}
0.2 = \dfrac{{{R_e}}}{{{R_e} + h}}\\
0.2h = 0.8{R_e}\\
h = 4 \times \left( {6400\;{\rm{km}}} \right)\\
h = 25,600\;{\rm{km}}
\end{array}\]
Therefore, the value of gravitational acceleration will become 4 % at a distance of 25,600 km from the earth surface.
Note:
In order to resolve the given problem, try to remember the concept of variation of the gravitational acceleration with the height above the surface of the earth. Moreover, it is to be remembered that the magnitude of gravitational acceleration is different for the different places on the earth. The impact of gravitational force and gravity is necessary to take into consideration in order to achieve the desired result for any given set of conditions.
Complete step by step answer:
The Acceleration due to gravity at height h is given as,
\[{g_1} = g{\left( {\dfrac{{{R_e}}}{{{R_e} + h}}} \right)^2}..............................\left( 1 \right)\]
Here, \[g\] is the gravitational acceleration on the surface of earth and \[{R_e}\] is the radius of earth.
As the gravitational acceleration is 4 % of the value on the surface of the earth. Then the value is given as,
\[{g_1} = \dfrac{{4g}}{{100}}\]
Substituting the value in equation 1 as,
\[\begin{array}{l}
{g_1} = g{\left( {\dfrac{{{R_e}}}{{{R_e} + h}}} \right)^2}\\
\left( {\dfrac{{4g}}{{100}}} \right) = g{\left( {\dfrac{{{R_e}}}{{{R_e} + h}}} \right)^2}\\
0.04 = {\left( {\dfrac{{{R_e}}}{{{R_e} + h}}} \right)^2}\\
0.2 = \dfrac{{{R_e}}}{{{R_e} + h}}
\end{array}\]
Further solving the above equation as,
\[\begin{array}{l}
0.2 = \dfrac{{{R_e}}}{{{R_e} + h}}\\
0.2h = 0.8{R_e}\\
h = 4 \times \left( {6400\;{\rm{km}}} \right)\\
h = 25,600\;{\rm{km}}
\end{array}\]
Therefore, the value of gravitational acceleration will become 4 % at a distance of 25,600 km from the earth surface.
Note:
In order to resolve the given problem, try to remember the concept of variation of the gravitational acceleration with the height above the surface of the earth. Moreover, it is to be remembered that the magnitude of gravitational acceleration is different for the different places on the earth. The impact of gravitational force and gravity is necessary to take into consideration in order to achieve the desired result for any given set of conditions.
Recently Updated Pages
Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Trending doubts
Difference Between Prokaryotic Cells and Eukaryotic Cells

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Net gain of ATP in glycolysis a 6 b 2 c 4 d 8 class 11 biology CBSE

Give two reasons to justify a Water at room temperature class 11 chemistry CBSE
