Eccentricity of elliptical path of earth

[Amith2006]

Sir,
The eccentricity of Earth's 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 Earth's 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 r_ap and r_per, 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 Earth's 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

 

[andrevdh]

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

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

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