The maximum and minimum speeds of a comet that orbits the Sun are 80 a...
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.
The maximum and minimum speeds of a comet that orbits the Sun are 80 a...
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