1. Compute for the value of the acceleration due to gravity of an object at an altitude equal to twice the radius of the Earth? (radius of Earth = 6.4 x 10^6 m)

6. A synchronous orbit is an orbit about a planet's equator that always remains above the same spot of the planet because the synchronous orbit matches the planet's rotational period. The pilot of the Mars spacecraft (which was named Cheetah 1 and had a 9556 kg mass) asked Tarzan to calculate the altitude of a synchronous orbit so they could remain over a particular geological feature on the Martian equator. Calculate the altitude above the Martian equator using the data above and Johannes Kepler's genius.

Two planets orbit a star. You can ignore the gravitational interactions between the planets. Planet 1 has orbital radius r₁, and planet 2 has r₂ = 4r₁ Planet 1 orbits with period T₁ Planet 2 orbits with period A. T₂ = -1/T₁ B. T₂=2T₁ C. T₂ = 4T₁ D. T₂=8T₁

Can you show me how to solve this?

The orbit of a planet around the sun is : O A. hyperbolic. O B. parabolic. OC. elliptic. OD. circular.

According to Lunar Laser Ranging experiments the average distance LM from the Earth to the Moon is approximately 3.85 × 10° km. The Moon orbits the Earth and completes one revolution in approximately 27.5 days (a sidereal month). How can the mass of the Earth be calculated using the information above? Select the correct statements. Select one or more: O a. Use Newton's third law. O b. Use Newton's first law. c. Use Coulombs law. O d. Use Newton's second law.

According to Lunar Laser Ranging experiment the average distance LM from the Earth to the Moon is approximately 3.92 x 105 km. The Moon orbits the Earth and completes one revolution relative to the stars in approximately 27.5 days (a sidereal month). Calculate the orbital velocity of the Moon in m/s. Answer: Choose...

Leave the moon's amplitude (semi-major axis) constant. How does the changing of the moon's orbital period change the calculated value of the planet's mass? a. Increasing the orbital period results in a higher calculated value for the mass of the planet. Decreasing the orbital period results in a lower calculated value for the mass of the planet. b. Increasing the orbital period results in a lower calculated value for the mass of the planet. Decreasing the orbital period results in a higher calculated value for the mass of the planet. c. Increasing the orbital period results in a higher calculates value for the mass of the planet. Decreasing the orbital period results in a higher calculated value for the mass of the planet.

2. The planet Jupiter has an orbital period of 4332.8 days, and its greatest distance (aphelion distance) from the Sun is 5.455 AU {where 1 AU = 1 Astronomical Unit = 1.496x10'' m}. The mass of the planet Jupiter is 1.898x1027 kg and the mass of the Sun is 1.988x1030 kg. {G = 6.67x10'" Nm/kg?) (a) Calculate the perihelion distance (closest distance) in Astronomical Units (AU) of the planet Jupiter from the Sun. (b) Calculate the eccentricity of planet Jupiter's orbit around the Sun. (c) Calculate the smallest speed of planet Jupiter around the Sun in km/s. At what point in Jupiter's orbit does Jupiter have its smallest speed?

010: A new planet (tentatively named "Melmac") is found in a circular orbit with a period of 571 years. The sun has a mass of 1.9891x1030 kg. How far away is the planet in Astronomical Units (AU) ? Note: An A.U. is 1.496x1011 m.