# An asteroid has a perihelion distance of 2.4 AU and an aphelion distance of 4.4 AU? How do you calculate its orbital semimajor axis, eccentricity and period?

a(1 - e) = 2.4 and a(1 + e) = 4.4. Solving, a = 3.4 AU and e =10/17 = 05882, nearly. The period is a X sqrt (a) = 6.27 years, nearly, using Kepler;s third law for approximation of the period..

Here, a is defined as 1 AU for the Earth, and when a is in AU units and period is expressed in years, period = a X sqrt (a).

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The orbital semi-major axis (a) can be computed as follows: a = (2.4 AU + 4.4 AU) / 2 = 3.4 AU. The eccentricity (e) of the orbit can be computed as follows: e = (aphelion distance - perihelion distance) / (aphelion distance + perihelion distance) = (4.4 AU - 2.4 AU) / (4.4 AU + 2.4 AU) = 0.5. The period (T) of the orbit can be computed using Kepler's third law: T^2 = a^3, where T is in years and a is in astronomical units (AU). Therefore, T^2 = (3.4 AU)^3 = 39.304 AU^3. Taking the square root of both sides yields T ≈ 7.485 years.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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