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The “New” Astronomy

The “New” Astronomy. What is the force of gravity on a satellite that is one Earth radius above the surface of the Earth? The weight of the satellite on Earth is 10,000 N?. The “New” Astronomy.

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The “New” Astronomy

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  1. The “New” Astronomy What is the force of gravity on a satellite that is one Earth radius above the surface of the Earth? The weight of the satellite on Earth is 10,000 N?

  2. The “New” Astronomy What is the force of gravity on a satellite that is one Earth radius above the surface of the Earth? The weight of the satellite on Earth is 10,000 N? At one Earth radius above the surface of the Earth, h = RE. Therefore g = G Mp / (Rp + h)2 g = G Mp / (Rp + RE )2 g = G Mp / (2 Rp )2 g = (¼) G Mp / ( Rp )2 = (¼) 9.8 m/sec2 Therefore W = m (1/4) g = ¼ m g = ¼ (10,000 N) = 2,500 N

  3. The “New” Astronomy A comet has an orbit that just touches the Earth’s orbit at perihelion and just touches the orbit of Neptune at aphelion. What is the period of the comet’s orbit? Assume the orbit of the earth and Neptune are circular.

  4. The “New” Astronomy A comet has an orbit that just touches the Earth’s orbit at perihelion and just touches the orbit of Neptune at aphelion. What is the period (approximately) of the comet’s orbit? Assume the orbit of the earth and Neptune are circular. If the orbits are considered to be circular, then the distance to perihelion is approximately 1 au, and the distance to aphelion is approximately 30.07 au. Therefore, the length of the major axis is 31.07, and the length of the semimajor axis is 15.535 au. From Kepler’s Third Law 15.5353≈ 3749 = P2 Therefore, the Period = 61.2 years

  5. The “New” Astronomy What is the eccentricity of the comet’s orbit in the previous question? Distance to perihelion = 1 au = a (1 – e) Where a is the length of the semimajor axis and e is the eccentricity. Therefore, 1 = 15.535 (1 – e) → e = 0.936

  6. The “New” Astronomy A satellite is to be placed in orbit 2 Earth radii above the surface of the Earth. At what speed will the satellite have to move in order to establish a circular orbit at the height? F = F = G {MEarth msatellite / r2} = msatellite v2 / r (circular orbit) Therefore, with some algebra, V2 = G MEarth / r = G MEarth / 3REarth The values of G, MEarth, and REarth can be found in tables in the book.

  7. The “New” Astronomy Where is the center of mass of the Sun and Jupiter when Jupiter is at aphelion?

  8. The “New” Astronomy Where is the center of mass of the Sun and Jupiter when Jupiter is at aphelion? At aphelion, the distance between the Sun and Jupiter is a (1 + e) = 5.203 (1+0.048) = 5.453 au Using the center of the sun as a reference, Xcm = (MSun xSun + MJupiter xJupiter)/(MSun + MJupiter) = (317.8 MEarth 5.453 au)/(330000 MEarth + 317.8 MEarth = (1732.96/330317.8) = .00525 au

  9. The “New” Astronomy Where is the center of mass of the Sun and Jupiter when Jupiter is at aphelion? At aphelion, the distance between the Sun and Jupiter is a (1 + e) = 5.203 (1+0.048) = 5.453 au Using the center of the sun as a reference, Xcm = (MSun xSun + MJupiter xJupiter)/(MSun + MJupiter) = (317.8 MEarth 5.453 au)/(330000 MEarth + 317.8 MEarth = (1732.96/330317.8) = .00525 au

  10. The “New” Astronomy Where is the center of mass of the Sun and Jupiter when Jupiter is at aphelion? The center of mass of the Sun-Jupiter system is shifted 0.00525 au from the center of the sun. The radius of the sun is 0.00465 au.

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