R1=1.47*1011
R2=1.52*1011
Mass of the Earth=5.972*1024=ME
Mass of the Sun=1.989*1030=MS
Change in potential energy=(GMSME/R1)-(GMSME/R2)
[Put value of G,MS,ME,R1&R2]
Change in potential energy=1.765*1012J
Earth's distance from the Sun varies from 1.47*10^11 m at perihelion to 1.52*10^11 m at phelion...
The semimajor axis of Mars orbit is about 1.52 astronomical units (au), where an au is the Earth's average distance from the Sun, meaning the semimajor axis of Earth's orbit is 1 au. To go from Earth to Mars and use the least energy from rocket fuel, the orbit has a semimajor axis of 1.26 au and an eccentricity of about 0.21. Starting at Earth's orbit, to follow this path we give the spacecraft an orbital velocity of 40 km/s. ...
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?
3. Complete the following statements about Earth's orbital revolution: [1 The perihelion occurs on The aphelion Earth's distance from the sun is increasing from the months of occurs on to Earth's distance from the sun is decreasing from the months of to For each location, on how many days each year does the sun reach its highest point in the sky for that location? (Hint: Think about if/when the sun passes overhead as Earth progresses through the seasons.) [1] 8....
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...
Accidently uploaded the photo in a recent question and need help
solving this one.
3. Earth's orbit. The Earth, with mass m and angular momentum L, moves around the Sun in arn elliptic orbit of eccentricity e. The equation of trajectory is given by, r=p1+ecoso 1+e where p is the distance of the closest approach to the Sun (perihelion) (a) Find the two components of the velocity as functions ofQie, v,(9) and vo(9). (8 marks) (b) Prove that the angle...
6. Given that the distance to the Moon is 384000 km, and taking the Moon’s orbit around Earth to be circular, estimate the speed (in kilometers per second) at which the Moon orbits the Earth. 7. The baseline in Figure 0.19 in the textbook is 100 m and the angle at B is 60 degrees. Using the tangent function, calculate the distance from A to the tree Ch. 1 Discussion Questions 2. The benefit of our current knowledge lets us...
2) Planet Velocities and Energy (10 pts) We talked about how planet formation involves the collisions of bodies (planetesimals, embryos) leading to the growth (and heating) of a planet. Let's think about the velocities and energies involved here. a) The speed of a body in its orbit around the Sun is given by the equation V2= GM.[(2/r) - (1/a)] Here Vis the speed of the body in m/s, G is the gravitational constant, M. is the mass of the Sun...
1 What is the Earth's orbital speed as it orbits the Sun. Take the Earth/Sun distance to be 93 million miles (taking the Earth's orbit to be circular), the mass of the Earth to be 5.97 X 102 kg and the mass of the Sun to be 1.987 X 10*0 kg. Solve this problem using Newton's 2nd law directly 2. A 55.7 kg mass is lifted (at a constant velocity) from the ground to a vertical height of 1275 m....
Q5.7: An asteroid is in circular orbit around the Sun. Its distance from the Sun is 7.3 times the average distance of Earth from the Sun. The period of this asteroid is 20 Earth years A 7.3 Earth years 144 Earth years 53 Earth years 3.8 Earth years E
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