If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth ‘r’ below and a height ‘r’ above the earth surface is:

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Overview Physics Gravitation Physics Class 11th

If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth ‘r’ below and a height ‘r’ above the earth surface is:

Option 1 -

$1 + \frac{r}{R_{E}} + \frac{r^{2}}{R_{E}^{2}} + \frac{r^{3}}{R_{E}^{3}}$

Option 2 -

$1 - \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}$

Option 3 -

$1 + \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} - \frac{r^{3}}{R_{E}^{3}}$

Option 4 -

$1 + \frac{r}{R_{E}} - \frac{r^{2}}{R_{E}^{2}} + \frac{r^{3}}{R_{E}^{3}}$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

Vishal Baghel | Contributor-Level 10

2 months ago

Correct Option - 3

Detailed Solution:

Acceleration due to gravity at r distance above the surface = $\frac{G M}{{(R + r)}^{2}}$

Acceleration due to gravity at r distance below the surface = $\frac{G M}{R^{3}} (R - r)$

So, ratio = $\frac{(R - r) {(R + r)}^{2}}{R^{3}} = \frac{(R - r) (R^{2} + r^{2} + 2 R r)}{R^{3}} = 1 + \frac{r}{R} - \frac{r^{2}}{R^{2}} - \frac{r^{3}}{R^{3}}$

Acceleration due to gravity at r distance above the surface = <math> <mrow> <mfrac> <mrow> <mi>G</mi> <mi>M</mi> </mrow> <mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>R</mi> <mo>+</mo> <mi>r</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math>  <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1025" DrawAspect="Content" ObjectID="_1815995968"> </o:OLEObject></xml><! [endif]->Acceleration due to gravity at r distance below the surface = <math> <mrow> <mfrac> <mrow> <mi>G</mi> <mi>M</mi> </mrow> <mrow> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mrow> <mo> (</mo> <mrow> <mi>R</mi> <mo>−</mo> <mi>r</mi> </mrow> <mo>)</mo> </mrow> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1026" DrawAspect="Content" ObjectID="_1815995969"> </o:OLEObject></xml><! [endif]->So, ratio = <math> <mrow> <mfrac> <mrow> <mrow> <mo> (</mo> <mrow> <mi>R</mi> <mo>−</mo> <mi>r</mi> </mrow> <mo>)</mo> </mrow> <msup> <mrow> <mrow> <mo> (</mo> <mrow> <mi>R</mi> <mo>+</mo> <mi>r</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mfrac> <mrow> <mrow> <mo> (</mo> <mrow> <mi>R</mi> <mo>−</mo> <mi>r</mi> </mrow> <mo>)</mo> </mrow> <mrow> <mo> (</mo> <mrow> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> <mo>+</mo> <mn>2</mn> <mi>R</mi> <mi>r</mi> </mrow> <mo>)</mo> </mrow> </mrow> <mrow> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> <mo>=</mo> <mn>1</mn> <mo>+</mo> <mfrac> <mrow> <mi>r</mi> </mrow> <mrow> <mi>R</mi> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msup> </mrow> </mfrac> <mo>−</mo> <mfrac> <mrow> <msup> <mrow> <mi>r</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> <mrow> <msup> <mrow> <mi>R</mi> </mrow> <mrow> <mn>3</mn> </mrow> </msup> </mrow> </mfrac> </mrow> </math> <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1027" DrawAspect="Content" ObjectID="_1815995970"> </o:OLEObject></xml><! [endif]->

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If RE be the radius of Earth, then the ratio between the acceleration due to gravity at a depth ‘r’ below and a height ‘r’ above the earth surface is:

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