Calculating the Earth-Sun distance during perihelion

[nukeman]

Calculate the Earth-Sun distance during perihelion (at Earth’s closest approach) The Earth’s orbit has a semi-major axis of a = 1.496×108 km and
eccentricity of e = 0.017. Is Earth’s orbit far from circular? Explain.

The formula to be used is: rP = a(1 − e)

My answer is: 158.82 - Is that correct?

Am i missing something?

 

[cepheid]

Your answer is meaningless unless if you specify what UNITS it is in.

Under the assumption that you meant km, your answer is clearly wrong.

 

[sachinism]

http://www.windows2universe.org/kids_space/earth_galore.html

 

[nukeman]

Your answer is meaningless unless if you specify what UNITS it is in.

Under the assumption that you meant km, your answer is clearly wrong.

sorry, yes, its KM

Can anyone help me with this?

 

[cepheid]

sorry, yes, its KM

Can anyone help me with this?

Sure. Can you show us the steps you used to calculate the perihelion distance?

First you need to compute 1-e. Then you need to multiply the result of that by a (the semi-major axis of the orbit).

Here's a hint: in this example, e is a very small number, agreed? Therefore, 1-e should be fairly close to 1. If that's true, then a*(1-e) should be fairly close to a. The answer you get should be close to the length of the semi-major axis. In other words, the closest earth-sun distance does not deviate very much from the average earth-sun distance. The fact that the earth-sun distance doesn't change very much as it goes around its orbit suggests that the orbit does not not deviate too much from circularity. In other words, it is not very elliptical (remember that e = 0 would be a perfect circle, so e very small means close to circular).

 

[nukeman]

Calculate the Earth-Sun distance during perihelion (at Earth’s closest approach) The Earth’s orbit has a semi-major axis of a = 1.496×108 km and
eccentricity of e = 0.017. Is Earth’s orbit far from circular? Explain.

The formula to be used is: rP = a(1 − e)

Sure.

I went (1 - e) which would be 1 - 0.017 Correct?

Then I just went a x 0.017 = 2.74

??

 

[cepheid]

Calculate the Earth-Sun distance during perihelion (at Earth’s closest approach) The Earth’s orbit has a semi-major axis of a = 1.496×108 km and
eccentricity of e = 0.017. Is Earth’s orbit far from circular? Explain.

The formula to be used is: rP = a(1 − e)

Sure.

I went (1 - e) which would be 1 - 0.017 Correct?

Then I just went a x 0.017 = 2.74

??

Yes, (1 - e) = (1 - 0.017) = 0.983. That is correct.

It's the next part that doesn't make any sense. The formula is a*(1-e), but for some reason you have written down a*e. I don't understand why. Not only that, but the answer doesn't make sense either. Please get into the habit of including units in all of your calculation steps. It is crucial.

The question is, what is a * (1 - 0.017)? Plug in the value for a in km.

 

[cepheid]

Okay, I just realized what part of the problem is. In the value that was given for the semi-major axis:

a = 1.496×108 km​

You think that the second number there is "one hundred and eight", don't you? It is not. This:

a = 1.496×10⁸ km​

is what was actually meant. Here's a helpful tip for how to catch mistakes like this in the future. Whenever you see a physical quantity, ask yourself, "does this number make any sense?" A distance of 150 km to the sun is clearly nonsensical. A distance of 150 MILLION km, however, is perfectly reasonable ;)

 

[nukeman]

Ahhhh, nice catch!

So, let's try this again. Please correct me :)

(1 - e) = (1 - 0.017) = 0.983

Now, 1.496x10^8 = 149,600,000

149,600,000 x .983 = 147,056,800 ? Is that correct?

 

[cepheid]

Yeah, that looks ok to me now