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What is the velocity of Earth at perihelion and aphelion? How is this information calculated?

Answer 1

An alternative approach: Assume that the average velocity 29.7848 km/s is attained when r = a = 1.496 E+08 km. Then the formula v = 29.7848Xsqrt (2a/r -1) gives mini/max 29.22 km/s and 30.29 km/s.

r = a(1 - e) = 1.471 E+08 km at perihelion and r = a(1 + e) = 1.521 E+08 km at aphelion, where e = 0.01671.

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Answer 2

Josiah Adams

Earth's perihelion velocity is #30.28#km/s and its aphelion velocity is #29.3#km/s.

Using Newton's equation, the force due to gravity which the Sun exerts of the Earth is given by: #F=(GMm)/r^2# Where #G# is the gravitational constant, #M# is the mass of the Sun, #m# is the mass of the Earth and #r# is the distance between the centre of the Sun and the centre of the Earth.

The centripetal force required to keep Earth in orbit is given by: #F=(mv^2)/r# Where #v# is the orbital velocity.

Combining the two equations, dividing by #m# and multiplying by #r# gives: #v^2 = (GM)/r#

The value of #GM=1.327*10^11km^3s^(-2)#.

At perihelion the distance from the Sun to the Earth is #147,100,000 km#. Substituting the values into the equation gives #v=30kms^(-1)#.

At aphelion the distance from the Sun to the Earth is #152,100,000 km#. Substituting the values into the equation gives #v=29.5kms^(-1)#.

The actual values as calculated using the NASA DE430 ephemeris data are #30.28ms^(-1)# and #29.3kms^(-1)#.

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Answer 3

Asher Cohen

Earth's velocity, calculated using Kepler's laws and the conservation of angular momentum, is approximately 30.29 km/s at perihelion and 29.78 km/s at aphelion.

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Answer from HIX Tutor

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.