Earth’s orbit is an ellipse with eccentricity 0.0167. Thus, earth’s distance from the sun and speed as it moves around the sun varies from day to day. This means that the length of the solar day is not constant through the year. Assume that earth’s spin axis is normal to its orbital plane and find out the length of the shortest and the longest day. A day should be taken from noon to noon. Does this explain variation of length of the day during the year?

This is a Long Type Questions as classified in NCERT Exemplar

Explanation-let m be the mass of the earth v_p, v_a be the velocity of the earth at perigee and apogee respectively. Similarly w_p and w_a are angular velocities.

At perigee, $r_p^2{w}_p=r_a^2{w}_a$ at apogee

If a is the semimajor axis of the earth's orbit then r_p=a (1-e) and r_a=a (1+e)

$\frac{w_p}{w_a}={\left(\frac{1+e}{1-e}\right)}^{2}e$ = 0.00167

$\frac{w_p}{w_a}=1.0691$

Let w be the angular speed which is geometric mean of w_p and w_a and corresponding to mean solar day.

$\left(\frac{w_p}{w}\right)\left(\frac{w}{w_a}\right)$ =1.0691

If w corresponds to 1⁰ per day then w_p = 1.034⁰ per day and w_a= 0.967⁰ per day . since 361⁰=24, mean solar day we get 361.034⁰ which corresponds to 24h,8.14' and 360.967⁰ corresponds to 23h59min52'. So

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This is a Long Type Questions as classified in NCERT Exemplar

Explanation-let m be the mass of the earth v_p, v_a be the velocity of the earth at perigee and apogee respectively. Similarly w_p and w_a are angular velocities.

At perigee, $r_p^2{w}_p=r_a^2{w}_a$ at apogee

If a is the semimajor axis of the earth's orbit then r_p=a (1-e) and r_a=a (1+e)

$\frac{w_p}{w_a}={\left(\frac{1+e}{1-e}\right)}^{2}e$ = 0.00167

$\frac{w_p}{w_a}=1.0691$

Let w be the angular speed which is geometric mean of w_p and w_a and corresponding to mean solar day.

$\left(\frac{w_p}{w}\right)\left(\frac{w}{w_a}\right)$ =1.0691

If w corresponds to 1⁰ per day then w_p = 1.034⁰ per day and w_a= 0.967⁰ per day . since 361⁰=24, mean solar day we get 361.034⁰ which corresponds to 24h,8.14' and 360.967⁰ corresponds to 23h59min52'. So this does not explain the actual variation of the length of the day during the year.

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<p>This is a Long Type Questions as classified in NCERT Exemplar</p><p><strong>Explanation-</strong>let m be the mass of the earth v_p, v_a be the velocity of the earth at perigee and apogee respectively. Similarly w_p and w_a are angular velocities.</p><div><div><picture><source srcset="https://images.shiksha.com/mediadata/images/articles/1745474655phpVkx2hb_480x360.jpeg" media=" (max-width: 500px)"><img src="https://images.shiksha.com/mediadata/images/articles/1745474655phpVkx2hb.jpeg" alt="" width="207" height="102"></picture></div></div><p>At perigee, <span contenteditable="false"> <math> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> <mo>=</mo> <msubsup> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> </msub> </math> </span>at apogee</p><p>If a is the semimajor axis of the earth's orbit then r_p=a (1-e) and r_a=a (1+e)</p><p><span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <msup> <mrow> <mrow> <mo> (</mo> <mfrac> <mrow> <mrow> <mn>1</mn> <mo>+</mo> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>-</mo> <mi>e</mi> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mi>e</mi> <mi> </mi> </math> </span>= 0.00167</p><p><span contenteditable="false"> <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> <mo>=</mo> <mn>1.0691</mn> </math> </span></p><p>Let w be the angular speed which is geometric mean of w_p and w_a and corresponding to mean solar day.</p><p><span contenteditable="false"> <math> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>w</mi> </mrow> </mrow> <mrow> <mrow> <mi>a</mi> </mrow> </mrow> </msub> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> </span>=1.0691</p><p>If w corresponds to 1⁰ per day then w_p = 1.034⁰ per day and w_a= 0.967⁰ per day . since 361⁰=24, mean solar day we get 361.034⁰ which corresponds to 24h,8.14' and 360.967⁰ corresponds to 23h59min52'. So this does not explain the actual variation of the length of the day during the year.</p>

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