Eccentricity of elliptical path of earth

[Amith2006]

Sir,
The eccentricity of earths orbit is 0.0167. What is the ratio of its maximum speed to its minimum speed in its orbit?
I solved it in the following way:
Let its maximum and minimum speed be v1 and v2 respectively. Let a and b be the semi length of the major and minor axis respectively. Let e be its eccentricity.
v is inversely proportional to a and b. I took this assumption because at points closest to the centre of the elliptical path the velocity is maximum.
Hence,
v1/v2 = a/b
Now for an ellipse, (1 – e^2) = (b/a)^2
By solving I get,
a/b = 1.00014
Therefore,
v1/v2 = 1.00014
But the book answer is 1.033. Is there any mistake?

[Janus]

Sir,
The eccentricity of earths orbit is 0.0167. What is the ratio of its maximum speed to its minimum speed in its orbit?
I solved it in the following way:
Let its maximum and minimum speed be v1 and v2 respectively. Let a and b be the semi length of the major and minor axis respectively. Let e be its eccentricity.
v is inversely proportional to a and b. I took this assumption because at points closest to the centre of the elliptical path the velocity is maximum.
Hence,
v1/v2 = a/b
Now for an ellipse, (1 – e^2) = (b/a)^2
By solving I get,
a/b = 1.00014
Therefore,
v1/v2 = 1.00014
But the book answer is 1.033. Is there any mistake?


Your mistake is in assuming that the v1 and V2 are porportional to a and b. They are instead proportional to rap and rper, which are the aphelion and perhelion of the orbit, the furthest distance and closest approach the orbit has to the Sun. (these are not equal to a and b. Since the Sun is located at one of the foci of the elipse and not the center.)

The aphelion is found by
r_{ap}= a(1+e)
and the perhelion by
r_{per}= a(1-e)

[Andrew Mason]

Sir,
The eccentricity of earths orbit is 0.0167. What is the ratio of its maximum speed to its minimum speed in its orbit?
I solved it in the following way:
Let its maximum and minimum speed be v1 and v2 respectively. Let a and b be the semi length of the major and minor axis respectively. Let e be its eccentricity.
v is inversely proportional to a and b. I took this assumption because at points closest to the centre of the elliptical path the velocity is maximum.
Hence,
v1/v2 = a/b
Now for an ellipse, (1 – e^2) = (b/a)^2
By solving I get,
a/b = 1.00014
Therefore,
v1/v2 = 1.00014
But the book answer is 1.033. Is there any mistake?
But a/b is not the ratio of maximum to minimum radii. A planet prescribes an orbit about the sun with the sun at one of the focii. So the ratio of max radius to min radius is: (1+e)/(1-e), which is 1.0340

Since angular momentum is conserved:

mvr = constant

Therefore: v \propto 1/r

So this is also the ratio of maximum to minimum speeds.

AM

[maverick280857]

You can prove yourself that

V_{max} = \sqrt{\frac{GM}{a}\left(\frac{1+e}{1-e}\right)}

V_{min} = \sqrt{\frac{GM}{a}\left(\frac{1-e}{1+e}\right)}

This would be a good exercise in Kepler's Laws and elliptic orbits for you :smile:

[andrevdh]

Andrew's approach to your question can also be derived from Kepler's second law as follows:
\frac{dA_{ap}}{dt}=\frac{dA_{per}}{dt}

\frac{r_{ap}ds_{ap}}{2dt}=\frac{r_{per}ds_{per}}{2 dt}

r_{ap}v_{ap}=r_{per}v_{per}