Solve the equation of motion F=-mg-km|v|2sin2(theta
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Solve the equation of motion F=-mg-km|v|2sin2(theta).
It is required to find the solution for the equation of motion:
It is known that the force is calculated as:
Step by step
Solved in 3 steps
- We will use differential equations to model the orbits and locations of Earth, Mars, and the spacecraft using Newton’s two laws mentioned above. Newton’s second law of motion in vector form is: F^→=ma^→ (1) where F^→ is the force vector in N (Newtons), and a^→ is the acceleration vector in m/s^2,and m is the mass in kg. Newton’s law of gravitation in vector form is: F^→=GMm/lr^→l*r^→/lr^→l where G=6.67x10^-11 m^3/s^2*kg is the universal gravitational constant, M is the mass of the larger object (the Sun), and is 2x10^30 kg, and m is the mass the smaller one (the planets or the spacecraft). The vector r^→ is the vector connecting the Sun to the orbiting objects. Step one ) The motion force in Equation(1), and the gravitational force in Equation(2) are equal. Equate the right hand sides of equations (1) and (2), and cancel the common factor on the left and right sides. Answer: f^→=ma^→ f=Gmm/lr^→l^2 a^→=Gmm/lr^→l^2 x r^→/lr^→l r^→=r^→/lr^→l * Gmm Could you please…Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is v| = 225 km/s and the orbital period of each is 11.6 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 x 1030 kg.) M XCM M Part 1 of 3 - Conceptualize From the given data, it is difficult to estimate a reasonable answer to this problem without working through the details and actually solving it. A reasonable guess might be that each star has a mass equal to or slightly larger than our Sun because fourteen days is short compared to the periods of all the Sun's planets. Part 2 of 3 - Categorize The only force acting on each star is the central gravitational force of attraction which results in a centripetal acceleration. When we solve Newton's second law, we can find the unknown mass in terms of the variables…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). Select the correct expression to calculate the velocity of the Moon. Select one: L'M O a. UM 2nLM O b. UM Тм O c. UM LM Тм 2nLM TM O d. UM
- Astronomical Datat Mass (kg) 1.99 × 1030 7.35 × 1022 3.30 × 1023 4.87 X 1024 5.97 X 1024 6.42 × 1023 1.90 × 1027 5.68 X 1026 8.68 x 1025 1.02 × 1026 1.31 × 1022 Body Sun Moon Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto 8.43 x 10^24 N 2.07 x 10^10 N O 843 N Radius (m) 6.96 × 108 1.74 X 106 2.44 X 106 6.05 × 106 6.37 X 106 3.39 X 106 6.99 X 107 5.82 X 107 2.54 x 107 2.46 X 107 1.15 X 106 What is the weight of a 75 kg astronaut on the surface of Neptune? O 2.07 x 10^5 N Orbit radius (m) 3.84 × 108 5.79 × 1010 1.08 × 10¹1 1.50 × 10¹1 2.28 × 10¹1 7.78 x 10¹1 1.43 × 10¹2 2.87 × 10¹2 4.50 × 1012 5.91 X 1012 Orbital period 27.3 d 88.0 d 224.7 d 365.3 d 687.0 d 11.86 y 29.45 y 84.02 y 164.8 y 247.9 yPlaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is V = 210 km/s and the orbital period of each is 11.5 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 x 1030 kg.) solar masses M XCM MAfter landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 46.0 cm. The explorer finds that the pendulum completes 98.0 full swing cycles in a time of 145 s. What is the magnitude of the gravitational acceleration on this planet? Express your answer in meters per second per secondgPlanet=(?)m/s^2
- express the speed of light (c = 2.9979 x 10^8 m/s) in km/year (take note that there is a leap year every 4 years, i.e. an additional day is added to the total calendar days every four years)One full year is 365.26 days. Using this number and the distance between the sun and the earth, calculate the Earth’s velocity. Question 8 options: 21.4 km/s 23.7 km/s 26.2 km/s 29.8 km/s 31.9 km/sTaking the age of Earth to be about 4 ✕ 109 years and assuming its orbital radius of 1.5 ✕ 1011 m has not changed and is circular, calculate the approximate total distance Earth has traveled since its birth (in a frame of reference stationary with respect to the Sun). ..............m Answer and simplify
- Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This statement implies that the masses of the two stars are equal (see figure below). Assume the orbital speed of each star is |v| 190 km/s and the orbital period of each is 12.9 days. Find the mass M of each star. (For comparison, the mass of our Sun is 1.99 x 1030 kg.) solar masses M XCM MProxima Centauri is part of the Alpha Centauri star system. It is the nearest star to the Solar system. Using the most advanced rocket technology we have available today, roughly how much time would it take to send a spaceship to Proxima and get back (round - trip time)? [Note: a "spaceship" is a spacecraft with a crew.] Choose the most correct answer; none of these are precisely correct. Group of answer choices 4.5 billion years 150,000 years 750 years 4.2 years 75 light - years 4.2 light-years