Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon
1 answer:
Answer:
16.63min
Explanation:
The question is about the period of the comet in its orbit.
To find the period you can use one of the Kepler's law:
T: period
G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2
r: average distance = 1UA = 1.5*10^11m
M: mass of the sun = 1.99*10^30 kg
By replacing you obtain:
the comet takes around 16.63min
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Answer:
1.55 m
Explanation:
The potential produced by a point charge, is inversely proportional to the distance from the charge to the point where the potential is being calculated, as follows:
As it only depends from the distance r, we can conclude that if the potential is the same for any point to a distance r from the point charge, the equipotencial surface must be a sphere of radius r.
Replacing q = +1.7*10⁻⁸ C, and k = 9*10⁹ N*m²/C², and V, by 120 V and 54 V, we can find the distance from the charge, to the points where we are calculating the potential V, as follows:
The distance between both points, is just the difference between the radius of both spheres, as follows:
r₂ - r₁ = 1.55 m