# The nearest star to our solar …

The nearest star to our solar system is 4.29 light years, away how much is the distance in terms of parsecs? How many parallax would this star show when viewed from two location of earth six months apartin its orbit around the sun.??? Plz give this answer. Short and easy🙃🙂

Pooja Ranjna 1 month, 3 weeks ago

ANSWER Given: The distance of the star is,=4.29 ly We know that one light year is the distance travelled by light in one year. 1 light year = 3×108×365×24×60×60                     =94608×1011m Therefore, the distance in 4.29 ly can be written as: 4.29 ly=405868.32×1011 m We know that 1 Parsec is also the unit of distance and its value is: 1 parsec=3.08×1016m Now, to express the distance in terms of parsec. 4.29ly=3.08×1016405868.32×1011​=1.32 parsec The diameter of the Earth's orbit is, d=3×1011m The angle made by the star on the Earth's orbit can be given by:  θ=405868.32×10113×1011​=7.39×10−6rad But, the angle covered in 1sec=4.85×10−6rad So, the parallax in viwing the star at two different positions of the Eart's orbit: for, 7.39×10−6rad=4.85×10−67.39×10−6​=1.52"

Pooja Ranjna 1 month, 3 weeks ago

answr LOGIN JOIN NOW ￼ What would you like to ask? 11th Physics Units and Measurement Measurement of Length The nearest star to our sol... PHYSICS ￼Asked on October 15, 2019 bySweety Jêthî The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun ? ￼HARD Share Save ANSWER Given: The distance of the star is,=4.29 ly We know that one light year is the distance travelled by light in one year. 1 light year = 3×108×365×24×60×60                     =94608×1011m Therefore, the distance in 4.29 ly can be written as: 4.29 ly=405868.32×1011 m We know that 1 Parsec is also the unit of distance and its value is: 1 parsec=3.08×1016m Now, to express the distance in terms of parsec. 4.29ly=3.08×1016405868.32×1011​=1.32 parsec The diameter of the Earth's orbit is, d=3×1011m The angle made by the star on the Earth's orbit can be given by:  θ=405868.32×10113×1011​=7.39×10−6rad But, the angle covered in 1sec=4.85×10−6rad So, the parallax in viwing the star at two different positions of the Eart's orbit: for, 7.39×10−6rad=4.85×10−67.39×10−6​=1.52"

Divyanshi Goswami 1 month, 3 weeks ago

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