At what depth below the surface of earth the acceleration due to gravity g will be half of its value 1600 km above the surface of Earth radius of earth 6400 km?

• Correct Answer: A

  Solution :

      Radius of earth R = 6400 km \ \[h=\frac{R}{4}\] Acceleration due to gravity at a height h \[{{g}_{h}}=g{{\left( \frac{R}{R+h} \right)}^{2}}\]\[=g{{\left( \frac{R}{R+\frac{R}{4}} \right)}^{2}}\]\[=\frac{16}{25}g\] At depth 'd' value of acceleration due to gravity             \[{{g}_{d}}=\frac{1}{2}{{g}_{h}}\]          (According to problem) Þ \[{{g}_{d}}=\frac{1}{2}\left( \frac{16}{25} \right)g\]\[\Rightarrow g\left( 1-\frac{d}{R} \right)\]\[=\frac{1}{2}\left( \frac{16}{25} \right)\ g\] By solving we get\[d=4.3\times {{10}^{6}}m\]

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Page 2

• Correct Answer: A

  Solution :

                  \[{g}'=g-{{\omega }^{2}}R{{\cos }^{2}}\lambda \] For weightlessness at equator \[\lambda =0{}^\circ \] and \[g'=0\] \ \[0=g-{{\omega }^{2}}R\]Þ \[\omega =\sqrt{\frac{g}{R}}=\frac{1}{800}\frac{rad}{\sec }\]

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Page 3

• Correct Answer: B

  Solution :

                  Weight on surface of earth, \[mg=500\,N\] and weight below the surface of earth at \[d=\frac{R}{2}\]             \[m{g}'=mg\left( 1-\frac{d}{R} \right)=mg\left( 1-\frac{1}{2} \right)=\frac{mg}{2}=250\,N\]

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Page 4

• Correct Answer: A

  Solution :

      \[\ g=\frac{4}{3}\pi GR\rho \] and \[{g}'=\frac{4}{3}\pi G{R}'\rho \] \ \[\frac{{{g}'}}{g}=\frac{{{R}'}}{R}=0.2\] Þ \[{g}'=0.2\,g\]

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Page 5

• Correct Answer: A

  Solution :

     

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Page 6

• Correct Answer: D

  Solution :

       \[\frac{{{g}_{m}}}{{{g}_{e}}}=\frac{{{M}_{m}}}{{{M}_{e}}}\times {{\left( \frac{{{R}_{e}}}{{{R}_{m}}} \right)}^{2}}=\left( \frac{1}{9} \right)\ {{\left( \frac{2}{1} \right)}^{2}}=\frac{4}{9}\]Þ \[{{g}_{m}}=\frac{4}{9}{{g}_{e}}\] \ \[{{W}_{\text{m}}}=\frac{4}{9}\times {{W}_{e}}=\frac{4}{9}\times 90=40\ kg\]

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Page 7

• Correct Answer: A

  Solution :

                  \[g'=g-{{\omega }^{2}}R,\] when w increases g' decreases.           

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Page 8

• Correct Answer: B

  Solution :

                  \[\frac{{{g}'}}{g}=\frac{{{M}'}}{M}\times \frac{{{R}^{2}}}{{{{{R}'}}^{2}}}=\frac{1}{2}\times \frac{4}{1}=\frac{2}{1}\]

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Page 9

• Correct Answer: B

  Solution :

                  Acceleration due to gravity at latitude l is given by             \[{g}'=g-R{{\omega }^{2}}{{\cos }^{2}}\lambda \] At \[{{30}^{o}},\,\,{{g}_{30{}^\circ }}=g-R{{\omega }^{2}}{{\cos }^{2}}{{30}^{o}}=g-\frac{3}{4}R{{\omega }^{2}}\] \ \[g-{{g}_{30}}=\frac{3}{4}{{\omega }^{2}}R.\]

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Page 10

• Correct Answer: B

  Solution :

                  Acceleration due to gravity g = \[\frac{GM}{{{R}^{2}}}\] \ \[\frac{g}{G}=\frac{M}{{{R}^{2}}}\]

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At what depth below the surface of the earth will acceleration due to gravity g be half its value 1600 km above the surface of the earth?

• 4.35 × 106 m

• 3.19 × 106 m

• 1.59 × 106 m

• 4.35 × 105 m

4.35 × 106 m

Explanation:

Radius of earth, R = 6400 km

∴ h = `R/4`

Gravitational acceleration at a height h,

gh = `g(R/(R + h))^2 = g(R/(R + R/4))^2 = 16/25` g

At depth 'd,' the value of gravity's acceleration

gd = `1/2` gh   ......(Given)

∴ gd = `1/2(16/25)g` 

∴ `g(1 - d/R) = 1/2(16/25)g`

∴ `1 - 16/50 = d/R`

∴ d = `(34 xx 6400 xx 10^3)/50`

∴ d = 4.35 × 106 m

Concept: Acceleration Due to Gravity Below and Above the Earth's Surface

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