Somewhere far, far away on the other side of our galaxy, there is a new protoplanetary disk where two baby planetesimals are growing. Planetesimal A is currently 5x10° kg, and Planetesimal B is 2x105 kg. If these two planetesimals are 100 meters apart, what is the gravitational force between them? (Hint: the Universal Gravitational Constant (G) is 6.67x10-11 Nm²/kg²) Gm1m2 FG %D r2 Planetesimal B Planetesimal A 2 x 10 kg 5 x 10°kg 100 meters
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- You have negotiated with the Omicronians for a base on the planet Omicron Persei 7. The architects working with you to plan the base need to know the acceleration of a freely falling object at the surface of the planet in order to adequately design the structures. The Omicronians have told you that the value is gOP7=7.29 flurg/grom^2, but your architects use the units meters/second^2, and from your previous experience you know that both the Omicronians and your architects are terrible at unit conversion. Thus, it's up to you to do the unit conversion. Fortunately, you know the unit equality relationships: 5.24flurg=1meter and 1grom=0.493second. What is the value of gOP7 in the units your architects will use, in meter per second squared?Since 1995, hundreds of extrasolar planets have been discovered. There is the exciting possibility that there is life on one or more of these planets. To support life similar to that on the Earth, the planet must have liquid water. For an Earth-like planet orbiting a star like the Sun, this requirement means that the planet must be within a habitable zone of 0.9 AU to 1.4 AU from the star. The semimajor axis of an extrasolar planet is inferred from its period. What range in periods corresponds to the habitable zone for an Earth-like Planet orbiting a Sun-like star?You are a rover pilot on the crew of the initial exploration team sent to Kepler 22b,the first extrasolar planet discovered within the habitable zone of a sun-like star. Thescience team recently discovered liquid water on the surface. (Hurrah!) Your rover isat point A on the shore of a circular lake with radius 4 km collecting samples. Thescience team wants to send your rover to a point C diametrically opposite A. Therover can drive around the circumference of the lake at a rate of 4 km per hour andfly over the lake at a rate of 3 km per hour.(a) How long will it take the rover to fly across the lake?(b) How long will it take the rover to drive around the shore of the lake?You could also fly at an angle θ along a chord inside the circular lake, andcomplete the rest of the path driving along the circumference of the lake.(c) Find the length of the chord in terms of θ. How long will it take the drone totraverse the chord?(d) Find the length of the remaining shoreline after the cord in…
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- Please answer the question and subquestions entirely. This is one single question. According to the official guideline, I can ask two subquestions! Thank you! 1) A planet Y is moving in circular orbit around the Sun. If its distance from the Sun is four times the average distance of the Earth from the Sun, what is the Y’s period in Earth years? 3 8 16 32 64 a) Two masses are precisely 1 m apart from each other. The gravitational force each exerts on the other is exactly 1 N. If the masses are identical, what is each mass? 1.22 x 105 kg 1.34 x 1010 kg 2.50 x 105 kg 1.58 x 1010 kg b) What is the acceleration due to gravity on the surface of the planet Pluto if its mass is 1.2 x 10 22 kg and radius is 1.14 x 10 6 m? 9.8 m/s2 6.4 m/s2 0.62 m/s2 0.34 m/s21. Consider our Sun - it is in orbit around the center of our Milky Way Galaxy. The velocity of the Sun in its orbit is about 250 km/s. The distance to the center of the galaxy is about 9.1 kpc (kiloparsecs). We can use Kepler's third law to calculate the mass of the galaxy interior to the Sun's orbit. We assume that the orbit is circular so that the semimajor axis is just the radius of the circular orbit = 9.1 kpc. First we need to calculate the number of AU's in 9.1 kpc. (Note that 1 Крс - 1000 рс - 3260 1t yrs and 1 pc - 206,265 AU.) %3D a =r =9.1kpc = (9.1kpc) 1000 pc 206,265AU] 1kpc AU Sun 1pcNeptune orbits the Sun with an orbital radius of 4.495 x 10^12 m. If the earth to sun distance 1A.U. = 1.5 x 10^11 m, a) Determine how many A.U.'s is Neptune's orbital radius (Round to the nearest tenth). b) Given the Sun's mass is 1.99 x10^30 kg, use Newton's modified version of Kepler's formula T^2 = (4pi^2/Gm(star)) x d^3 to find the period in seconds using scientific notation. (Round to the nearest thousandth). C) Convert the period in part b) to years (Round to the nearest tenth)
- How hard do two planets pull on each other if their masses are 150,000 kg and 200,000 kg and they 230 million kilometers apart? O 5 x 10^-23 N O 3.78 x 10^-23 N 2.5 x 10^-23N O 4.23 x 10^-23 NProblem Solving INSTRUCTION: When doing a computation for each questionnaire, round-off your FINAL ANSWER ONLY to two decimal places (do not round off during computation). How hard do two planets pull on each other if their masses are 150,000 kg and 200,000 kg and they 230 million kilometers apart? 5 x 10^-23 N O 3.78 x 10^-23 N 2.5 x 10^-23 N O 4.23 x 10^-23 NE Native American mascots - hor x SI Course Modules: AST 111: Intro x A Ch 21: Venus and Mars - AST 1 x © Squaring both sides and solvin x| + A webassign.net/web/Student/Assignment-Responses/submit?pos=16&dep=24621113&tags=autosave#question4793215_16 Tutorial The Magellan orbiter orbits Venus with a period of 3.26 hours. How far (in km) above the surface of the planet is it? (The mass of Venus is 4.87 x 1024 kg, and the radius of Venus is 6.05 x 103 km.) Part 1 of 3 The period of the orbiter's orbit can give us the speed at which the orbiter orbits the planet. We imagine the orbiter tracing a circle around the planet at a certain height, the speed is 2ar V = P Part 2 of 3 Next, we combine this with the circular velocity equation to determine the height above the planet's surface. GM V = 2ar GM Squaring both sides and solving for r gives the following equation. What is the exponent for r? GM Part 3 of 3 Congratulations! You just derived a version of Kepler's Third Law for Venus! Using…