A particle balls towards earth from infinity. Its velocity on reaching the earth will be (R is redius of earth)
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Heya user☺☺
We can see this from the expression for the potential energy at a distance rr:
V=−GMmr
For the Sun M=1.9891×1030M=1.9891×1030 kilograms and rr (the orbital radius of the Earth) ≈1.5×1011≈1.5×1011 so V≈8.85×108mV≈8.85×108m J.
For the Earth M=5.97219×1024M=5.97219×1024 kilograms and rr (the radius of the Earth) ≈6.4×106≈6.4×106 so V≈6.15×107mV≈6.15×107m J.
So the effect of the Sun's gravity is about 14 times greater. The velocity of the comet will be given by:
12mv2=(8.85×108+6.15×107)m12mv2=(8.85×108+6.15×107)m
which gives:
v≈43.5km/sec
Hope this will help☺☺
We can see this from the expression for the potential energy at a distance rr:
V=−GMmr
For the Sun M=1.9891×1030M=1.9891×1030 kilograms and rr (the orbital radius of the Earth) ≈1.5×1011≈1.5×1011 so V≈8.85×108mV≈8.85×108m J.
For the Earth M=5.97219×1024M=5.97219×1024 kilograms and rr (the radius of the Earth) ≈6.4×106≈6.4×106 so V≈6.15×107mV≈6.15×107m J.
So the effect of the Sun's gravity is about 14 times greater. The velocity of the comet will be given by:
12mv2=(8.85×108+6.15×107)m12mv2=(8.85×108+6.15×107)m
which gives:
v≈43.5km/sec
Hope this will help☺☺
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