Solution)
Acc to Conservation of Energy
V2=sqrt(v1^2 + 2GM (1/d1 - 1/d2))
V2= sqrt((1*10^4)^2 + 2*(6.67*10^-11)*(1.99*10^30)(1/6*10^10 - 1/ 2.9*10^11))
V2= 1000 m/s
=======
Acc to HOMEWORKLIB RULES, I can answer only one at a time. Good luck!:)
comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed...
omets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.0×104 m/s when at a distance of 2.4×1011 m from the center of the sun, what is its speed when at a distance of 6.0×1010 m . Express your answer using two significant figures. v= ?m/s Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed 2.0 times 10^4 m/s when at a distance of 2.4 times 10^11...
A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 5 1010 m (inside the orbit of Mercury), at which point its speed is 9 104 m/s. Its farthest distance from the Sun is far beyond the orbit of Pluto. What is its speed when it is 6 1012 m from the Sun? (This is the approximate distance of Pluto from the Sun.) speed = ______m/s
8. A comet with unknown mass mc is in an elliptical orbit around the sun: R. Rp mc mc MS ū When the comet is at its parahelion R, it has a speed of up = 80km/s and at its aphelion R, it has a speed of va = 10km/s. For this problem suppose we do not know the mass of the sun M, or the value of Newton's constant G, but we can approximate the orbit of the earth...
A planet is discovered to orbit around a star in the galaxy Andromeda, with the same orbital diameter as the Earth around our Sun. If that star has 4 times the mass of our Sun, what will the period of revolution of that new planet be, compared to the Earth's orb o One-fourth as much o One-half as much Twice as much A. Four times much The average distance from the Earth to the Sun is defined as one "astronomical...
The average distance from the Earth to the Sun is defined as one "astronomical unit" (AU). An asteroid orbits the Sun in one-third of a year. What is the asteroids average distance from the Sun? 0.19 AU 0.43 AU o 21 AU 5.2 AU Many comets orbit the Sun in highly elliptical orbits. What happens over time to the comet's angular momentum? It continually increases. It continually decreases. It remains constant It increases during some parts of the orbit and...
A comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being about 0.620 AU and its greatest distance from the sun being 35.5 AU (1 AU = the Earth-Sun distance). If the comet's speed at closest approach is 54.0 km/s, what is its speed when it is farthest from the Sun? (The gravitational force exerted by the Sun on the comet is parallel to the moment arm, so exerts no torque. Therefore, angular...
(1) Halley's Comet and Kepler's Laws: Halley's Comet orbits the Sun every 76.0 years. Its elliptical orbit has an eccentricity of 0.97. Using that information and the Kepler material in Math Insight Box 3.2 (p.70) and Figures 3.16, 3.17 and 3.18: 40 points total] (a) Find the comet's average distance from the Sun (semimajor axis, a) (b) Find its closest and furthest distances from the Sun: the Perihelion & Aphelion distances.
The orbit of a 1.5 ✕ 1010 kg comet around the Sun is elliptical, with an aphelion distance of 33.0 AU and perihelion distance of 0.850 AU. (Note: 1 AU = one astronomical unit = the average distance from the Sun to the Earth = 1.496 ✕ 1011 m.) (a)What is its orbital eccentricity? (b)What is its period? (Enter your answer in yr.) (c)At aphelion what is the potential energy (in J) of the comet—Sun system?
A)Some comets are in highly elongated orbits that come very close to the sun at perihelion. The distance from one such comet to the center of the sun is 6.00×10^9m at perihelion and 3.00×10^12m at aphelion. For this comet's orbit, find the semi-major axis. B) In 2017 astronomers discovered a planet orbiting the star HATS-43. The orbit of the planet around HATS-43 has semi-major axis 7.41×109m, eccentricity 0.173, and period 4.39 days. Find the distance between HATS-43 and the planet...
(a) An asteroid is in an elliptical orbit around a distant star. At its closest approach, the asteroid is 0.700 AU from the star and has a speed of 54.0 km/s. When the asteroid is at its farthest distance from the star of 36.0 AU, what is its speed (in km/s)? (1 AU is the average distance from the Earth to the Sun and is equal to 1.496 x 101 m. You may assume that other planets and smaller objects...