Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

4 0

4 years ago

If the moon is rising at midnight, the phase of the moon must be

Nadya [2.5K]

The phase is called 3rd quarter.

Hope this helps:)

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3 years ago

If this resolving power is diffraction-limited, to what effective diameter of your eye's optical system does this correspond? us

ZanzabumX [31]

"This resolving power" was obviously stated earlier, somewhere before the point where you started copying. With no resolving power specified, there's actually no question, and so no answer.

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3 years ago

Does a five pound ball weigh more or less 40000 miles above sea level​

Does a five pound ball weigh more or less 40000 miles above sea level