The maximum and minımum distances of a comet from the Sun are 1.6 × 1012 m and 8.0 × 1010 m respectively. If the speed of the comet at the nearest point is 6 × 104 ms -1, the speed at the farthest point is:

Question

The maximum and minımum distances of a comet from the Sun are 1.6 × 1012 m and 8.0 × 1010 m respectively. If the speed of the comet at the nearest point is 6 × 104 ms -1, the speed at the farthest point is: (A) 1.5 × 103 m / s (B) 6.0 × 103 m / s (C) 3.0 × 103 m / s (D) 4.5 × 103 m / s

Tardigrade Admin · Accepted Answer

By angular momentum conservation: mv 1 r 1= mv 2 r 2 v 1=(48 × 1014/1.6 × 1012)=3000 m / sec =3 × 103 m / sec

Q. The maximum and minımum distances of a comet from the Sun are $1.6 \times 1 0^{12} m$ and $8.0 \times 1 0^{10} m$ respectively. If the speed of the comet at the nearest point is $6 \times 1 0^{4} m s^{- 1}$ , the speed at the farthest point is:

$1.5 \times 1 0^{3} m / s$

$6.0 \times 1 0^{3} m / s$

$3.0 \times 1 0^{3} m / s$

$4.5 \times 1 0^{3} m / s$

By angular momentum conservation :
$m v_{1} r_{1} = m v_{2} r_{2}$
$v_{1} = \frac{48 \times 1 0 ^{14}}{1.6 \times 1 0 ^{12}} = 3000 m / sec$
$= 3 \times 1 0^{3} m / sec$

Q. The maximum and minımum distances of a comet from the Sun are 1.6×1012m and 8.0×1010m respectively. If the speed of the comet at the nearest point is 6×104ms−1, the speed at the farthest point is:

1.5×103m/s

6.0×103m/s

3.0×103m/s

4.5×103m/s