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Question: The earth moves round the sun once in a year with a speed of 30 kilometre per second. What is the ce...

The earth moves round the sun once in a year with a speed of 30 kilometre per second. What is the centripetal acceleration of earth towards the sun ?
(A) 6×10−7m/s26 \times {10^{ - 7}}m/{s^2}
(B) 6×10−6m/s26 \times {10^{ - 6}}m/{s^2}
(C) 6×10−5m/s26 \times {10^{ - 5}}m/{s^2}
(D) 6×10−3m/s26 \times {10^{ - 3}}m/{s^2}

Explanation

Solution

In order to solve this problem, first we calculate the distance between earth and sun by following expression :
d=ctd = ct
Where
d == distance between earth and sun
c == speed of light
t == time taken by light from sun to reach earth
After then apply this value in the formula of centripetal acceleration i.e., ac=v2R{a_c} = \dfrac{{{v^2}}}{R}
Here R=d=R = d = distance between sun and earth
v == velocity of earth

Step by step answer:
Given that the velocity of earth when it revolves around the sun is 30 kilometre per second.
ve=3×104m/s{v_e} = 3 \times {10^4}m/s …..(1)
We know that light takes 8 minutes from the sun to reach earth.
t=8minutet = 8minute
t=8×60secondt = 8 \times 60 second
t=480secondst = 480 seconds
So, the distance between earth and sun is
d=c×td = c \times t
Here c=3×108m/sc = 3 \times {10^8}m/s ((speed of light in vacuum))
d=3×108×480d = 3 \times {10^8} \times 480
d=1440×108md = 1440 \times {10^8}m
d=144×109md = 144 \times {10^9}m …..(2)
Earth revolves around the sun in a circular orbit. So centripetal acceleration is given as
ac=ve2d{a_c} = \dfrac{{v_e^2}}{d}
On putting the values from equation 1 and 2
ac=(3×104)2144×109{a_c} = \dfrac{{{{(3 \times {{10}^4})}^2}}}{{144 \times {{10}^9}}}
ac=3×3×104×104144×109{a_c} = \dfrac{{3 \times 3 \times {{10}^4} \times {{10}^4}}}{{144 \times {{10}^9}}}
ac=9144×108×10−9{a_c} = \dfrac{9}{{144}} \times {10^8} \times {10^{ - 9}}
ac=9144×10−1{a_c} = \dfrac{9}{{144}} \times {10^{ - 1}}
ac=900144×10−3{a_c} = \dfrac{{900}}{{144}} \times {10^{ - 3}}
ac=6.25×10−3m/s2{a_c} = 6.25 \times {10^{ - 3}}m/{s^2}
Hence option D is correct answer i.e., 6×10−3m/s26 \times {10^{ - 3}}m/{s^2}

Note: in circular motion, the main reason for circular direction of a particle is centripetal acceleration. i.e., centripetal acceleration is the property of the motion of a body traversing a circular path and the direction is radially towards the centre of the circle.