359692
The value of gravitational acceleration \(g\) at a height \(h\) above the earth's surface is \(\dfrac{g}{4}\), then (\(R=\) radius of earth \()\)
359693
At what altitude \((h)\) above the earth's surface would the acceleration due to gravity be one fourth of its value at the earth's surface?
1 \(h=R\)
2 \(h=4 R\)
3 \(h=2 R\)
4 \(h=16 R\)
Explanation:
Given, \(g^{\prime}=\dfrac{g}{4}\) We know that, acceleration due to gravity at height \(h\) from the surface of the earth, \(g^{\prime}=g\left[\dfrac{R}{R+h}\right]^{2}\) Hence, \(\quad \dfrac{g}{4}=g\left[\dfrac{R}{R+h}\right]^{2}\left(\because g^{\prime}=\dfrac{g}{4}\right)\) \(\Rightarrow \dfrac{R}{R+h}=\dfrac{1}{2}\) \(\Rightarrow \quad R+h=2 R \Rightarrow h=R\) where, \(R\) is the radius of the earth.
PHXI08:GRAVITATION
359694
The acceleration due to gravity at height \(h\) above the earth if \(h \ll R\) (Radius of earth) is given by
The value of acceleration due to gravity at height ' \(h\) ' above the earth surface is given by, \(g^{\prime}=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^{2}}\) Using binomial expansion \(\cong g^{\prime}=g\left(1-\dfrac{2 h}{R}\right)(\) if \(h \ll R)\) So, correct option is (1).
JEE - 2023
PHXI08:GRAVITATION
359695
The acceleration due to gravity becomes \((g/2)\) (where \(g=\) acceleration due to gravity on the surface of the earth ) at a height equal to
1 \(4R\)
2 \(R/4\)
3 \(2R\)
4 \(R / 2\)
Explanation:
\(\begin{aligned}g & =\dfrac{G M}{R^{2}}, g^{\prime}=\dfrac{G M}{(R+h)^{2}} \\\therefore & \dfrac{g}{g^{\prime}}=\left(\dfrac{R+h}{R}\right)^{2}=\left(1+\dfrac{h}{R}\right)^{2} \\\Rightarrow & \dfrac{g^{\prime}}{g}=\left(1+\dfrac{h}{R}\right)^{-2}=\left(1-\dfrac{2 h}{R}\right), \text { when } h < < R . \\& \text { or } \dfrac{g / 2}{g}=1-\dfrac{2 h}{R} \Rightarrow h=\dfrac{R}{4}\end{aligned}\)
359692
The value of gravitational acceleration \(g\) at a height \(h\) above the earth's surface is \(\dfrac{g}{4}\), then (\(R=\) radius of earth \()\)
359693
At what altitude \((h)\) above the earth's surface would the acceleration due to gravity be one fourth of its value at the earth's surface?
1 \(h=R\)
2 \(h=4 R\)
3 \(h=2 R\)
4 \(h=16 R\)
Explanation:
Given, \(g^{\prime}=\dfrac{g}{4}\) We know that, acceleration due to gravity at height \(h\) from the surface of the earth, \(g^{\prime}=g\left[\dfrac{R}{R+h}\right]^{2}\) Hence, \(\quad \dfrac{g}{4}=g\left[\dfrac{R}{R+h}\right]^{2}\left(\because g^{\prime}=\dfrac{g}{4}\right)\) \(\Rightarrow \dfrac{R}{R+h}=\dfrac{1}{2}\) \(\Rightarrow \quad R+h=2 R \Rightarrow h=R\) where, \(R\) is the radius of the earth.
PHXI08:GRAVITATION
359694
The acceleration due to gravity at height \(h\) above the earth if \(h \ll R\) (Radius of earth) is given by
The value of acceleration due to gravity at height ' \(h\) ' above the earth surface is given by, \(g^{\prime}=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^{2}}\) Using binomial expansion \(\cong g^{\prime}=g\left(1-\dfrac{2 h}{R}\right)(\) if \(h \ll R)\) So, correct option is (1).
JEE - 2023
PHXI08:GRAVITATION
359695
The acceleration due to gravity becomes \((g/2)\) (where \(g=\) acceleration due to gravity on the surface of the earth ) at a height equal to
1 \(4R\)
2 \(R/4\)
3 \(2R\)
4 \(R / 2\)
Explanation:
\(\begin{aligned}g & =\dfrac{G M}{R^{2}}, g^{\prime}=\dfrac{G M}{(R+h)^{2}} \\\therefore & \dfrac{g}{g^{\prime}}=\left(\dfrac{R+h}{R}\right)^{2}=\left(1+\dfrac{h}{R}\right)^{2} \\\Rightarrow & \dfrac{g^{\prime}}{g}=\left(1+\dfrac{h}{R}\right)^{-2}=\left(1-\dfrac{2 h}{R}\right), \text { when } h < < R . \\& \text { or } \dfrac{g / 2}{g}=1-\dfrac{2 h}{R} \Rightarrow h=\dfrac{R}{4}\end{aligned}\)
359692
The value of gravitational acceleration \(g\) at a height \(h\) above the earth's surface is \(\dfrac{g}{4}\), then (\(R=\) radius of earth \()\)
359693
At what altitude \((h)\) above the earth's surface would the acceleration due to gravity be one fourth of its value at the earth's surface?
1 \(h=R\)
2 \(h=4 R\)
3 \(h=2 R\)
4 \(h=16 R\)
Explanation:
Given, \(g^{\prime}=\dfrac{g}{4}\) We know that, acceleration due to gravity at height \(h\) from the surface of the earth, \(g^{\prime}=g\left[\dfrac{R}{R+h}\right]^{2}\) Hence, \(\quad \dfrac{g}{4}=g\left[\dfrac{R}{R+h}\right]^{2}\left(\because g^{\prime}=\dfrac{g}{4}\right)\) \(\Rightarrow \dfrac{R}{R+h}=\dfrac{1}{2}\) \(\Rightarrow \quad R+h=2 R \Rightarrow h=R\) where, \(R\) is the radius of the earth.
PHXI08:GRAVITATION
359694
The acceleration due to gravity at height \(h\) above the earth if \(h \ll R\) (Radius of earth) is given by
The value of acceleration due to gravity at height ' \(h\) ' above the earth surface is given by, \(g^{\prime}=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^{2}}\) Using binomial expansion \(\cong g^{\prime}=g\left(1-\dfrac{2 h}{R}\right)(\) if \(h \ll R)\) So, correct option is (1).
JEE - 2023
PHXI08:GRAVITATION
359695
The acceleration due to gravity becomes \((g/2)\) (where \(g=\) acceleration due to gravity on the surface of the earth ) at a height equal to
1 \(4R\)
2 \(R/4\)
3 \(2R\)
4 \(R / 2\)
Explanation:
\(\begin{aligned}g & =\dfrac{G M}{R^{2}}, g^{\prime}=\dfrac{G M}{(R+h)^{2}} \\\therefore & \dfrac{g}{g^{\prime}}=\left(\dfrac{R+h}{R}\right)^{2}=\left(1+\dfrac{h}{R}\right)^{2} \\\Rightarrow & \dfrac{g^{\prime}}{g}=\left(1+\dfrac{h}{R}\right)^{-2}=\left(1-\dfrac{2 h}{R}\right), \text { when } h < < R . \\& \text { or } \dfrac{g / 2}{g}=1-\dfrac{2 h}{R} \Rightarrow h=\dfrac{R}{4}\end{aligned}\)
359692
The value of gravitational acceleration \(g\) at a height \(h\) above the earth's surface is \(\dfrac{g}{4}\), then (\(R=\) radius of earth \()\)
359693
At what altitude \((h)\) above the earth's surface would the acceleration due to gravity be one fourth of its value at the earth's surface?
1 \(h=R\)
2 \(h=4 R\)
3 \(h=2 R\)
4 \(h=16 R\)
Explanation:
Given, \(g^{\prime}=\dfrac{g}{4}\) We know that, acceleration due to gravity at height \(h\) from the surface of the earth, \(g^{\prime}=g\left[\dfrac{R}{R+h}\right]^{2}\) Hence, \(\quad \dfrac{g}{4}=g\left[\dfrac{R}{R+h}\right]^{2}\left(\because g^{\prime}=\dfrac{g}{4}\right)\) \(\Rightarrow \dfrac{R}{R+h}=\dfrac{1}{2}\) \(\Rightarrow \quad R+h=2 R \Rightarrow h=R\) where, \(R\) is the radius of the earth.
PHXI08:GRAVITATION
359694
The acceleration due to gravity at height \(h\) above the earth if \(h \ll R\) (Radius of earth) is given by
The value of acceleration due to gravity at height ' \(h\) ' above the earth surface is given by, \(g^{\prime}=\dfrac{g}{\left(1+\dfrac{h}{R}\right)^{2}}\) Using binomial expansion \(\cong g^{\prime}=g\left(1-\dfrac{2 h}{R}\right)(\) if \(h \ll R)\) So, correct option is (1).
JEE - 2023
PHXI08:GRAVITATION
359695
The acceleration due to gravity becomes \((g/2)\) (where \(g=\) acceleration due to gravity on the surface of the earth ) at a height equal to
1 \(4R\)
2 \(R/4\)
3 \(2R\)
4 \(R / 2\)
Explanation:
\(\begin{aligned}g & =\dfrac{G M}{R^{2}}, g^{\prime}=\dfrac{G M}{(R+h)^{2}} \\\therefore & \dfrac{g}{g^{\prime}}=\left(\dfrac{R+h}{R}\right)^{2}=\left(1+\dfrac{h}{R}\right)^{2} \\\Rightarrow & \dfrac{g^{\prime}}{g}=\left(1+\dfrac{h}{R}\right)^{-2}=\left(1-\dfrac{2 h}{R}\right), \text { when } h < < R . \\& \text { or } \dfrac{g / 2}{g}=1-\dfrac{2 h}{R} \Rightarrow h=\dfrac{R}{4}\end{aligned}\)
