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
Verified
474k+ 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
Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE
The highest possible oxidation states of Uranium and class 11 chemistry CBSE
Find the value of x if the mode of the following data class 11 maths CBSE
Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE
A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE
Statement I Reactivity of aluminium decreases when class 11 chemistry CBSE
Trending doubts
The reservoir of dam is called Govind Sagar A Jayakwadi class 11 social science CBSE
10 examples of friction in our daily life
Difference Between Prokaryotic Cells and Eukaryotic Cells
State and prove Bernoullis theorem class 11 physics CBSE
Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE
State the laws of reflection of light