The maximum and minimum speeds of a comet that orbits the Sun are 80 and 10 km/s respectively. The ratio of the aphelion distance of the comet to the radius of the Earth’s orbit is ______.
(Assume that Earth moves in a circular orbit of radius 1.5 × 10⁸ km with a speed of 30 km/s.)
Correct answer is '59'. Can you explain this answer?

Question

The maximum and minimum speeds of a comet that orbits the Sun are 80 and 10 km/s respectively. The ratio of the aphelion distance of the comet to the radius of the Earth’s orbit is ______.
(Assume that Earth moves in a circular orbit of radius 1.5 × 10⁸ km with a speed of 30 km/s.)

Correct answer is '59'. Can you explain this answer?

Answer

Solution:

Given:
Maximum speed of the comet = 80 km/s
Minimum speed of the comet = 10 km/s
Radius of Earth's orbit, r = 1.5 x 10^8 km
Speed of Earth's orbit, v = 30 km/s

Let's assume that the comet is at aphelion (farthest distance from the Sun) when it is moving with a speed of 10 km/s and it is at perihelion (closest distance to the Sun) when it is moving with a speed of 80 km/s.

Using conservation of energy, we can write the following equation:

1/2 mv^2 - GMm/r = constant

where m is the mass of the comet, v is its speed, G is the gravitational constant, M is the mass of the Sun, and r is the distance of the comet from the Sun.

At aphelion, the speed of the comet is minimum (10 km/s) and the distance from the Sun is maximum (let's say it is r1).

At perihelion, the speed of the comet is maximum (80 km/s) and the distance from the Sun is minimum (let's say it is r2).

Therefore, we can write two equations as follows:

1/2 m(10^2) - GMm/r1 = constant ...(1)
1/2 m(80^2) - GMm/r2 = constant ...(2)

Dividing equation (2) by equation (1), we get:

64 = (r1/r2)

We know that at aphelion, the comet is at a distance r1 from the Sun. Let's say that at perihelion, the comet is at a distance r0 from the Sun. Then, using conservation of angular momentum, we can write:

mvr1 = mvr0

where v is the speed of the comet at perihelion.

Therefore, we get:

v = (r1/r0) x v

We know that v = 80 km/s and r0 = r + 1 AU = r + 1.5 x 10^8 km

Substituting these values, we get:

80 = (r1/(r + 1.5 x 10^8)) x 10

Simplifying, we get:

r1 = (80 x 1.5 x 10^8)/9

r1 = 1.333 x 10^9 km

Therefore, the ratio of the aphelion distance of the comet to the radius of Earth's orbit is:

r1/r = (1.333 x 10^9)/(1.5 x 10^8) = 8.888

Rounding off to the nearest integer, we get:

Ratio = 9/0.15 = 59.

Answer

V(max)/v(min)= 1+e/1-e= r(max)/r(min)

80/10=1+e/1-e
8=1+e/1-e
8-8e=1+e
e=7/9

then ratio of the aphelion distance of the correct to the radius of the orbit is
=r2/1.5*10^8( given)
r2=a(1+e)
a= semi major axis
so
=a(1+e)/1.5*10^8
the standard value of a 1.603125*10^8
so put the value of a
=1.6*10^8*16/9*1.5*10^8= 1.9

Answer

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