In: Physics
The motion of Halley’s comet and its motion. Halley’s comet travels in an elliptical orbit of eccentricity ϵ = 0.97 around the Sun. At perihelion (closest approach), Halley’s comet is observed to be approximately 0.59 AU from the Sun. At aphelion the distance is about 35.08 AU, the semi-major axis of the elliptical orbit is 17.83 AU, and the orbital period is about 75.3 Earth years.
1) Since Earth has an essentially circular orbit that is 1 AU from the Sun. Use any approach that you like to determine the Earth’s orbital speed is v_E = sqrt(GM_S/R_E) and then determine a numerical value in kilometers/second.
2) Use the perihelion, aphelion, semi-major axis, and period above for Halley’s comet to determine the value of the characteristic length, r_c, that describes the elliptical orbital path, r(φ), for Halley’s comet.
3) Use the definition of r_c to estimate the speed of Halley’s comet at perihelion. Write the result in symbolic first, which should look like the result in part 1, then write it as the result from part (a) and appropriate ratios to estimate the numerical value of Halley’s comet’s speed at perihelion.
where is the gravitational constant. Equation (1) rearranges to give
Now, plugging in the numerical values,
, , , we get
Let the characteristic length of the elliptical orbit of Halley's comet be . Then, the semi-major axis length is given by
from which we can calculate as
Thus, the orbital position of the comet as a function of the angle is
where is the mass of Halley's comet. There is no need for subscript for angular momentum as it is a conserved quantity in central force motion. Now, the angular momentum is related to the characteristic length according to
Comparing (7) and (8), we get
from which the perihelion velocity becomes
which is the expression for the perihelion speed of the comet. Now, from the given data,
Thus, the perihelion speed is
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