A planet of mass m moves around a star of mass M in an elliptical orbit of semimajor axis a, and semi minor axis b. The total energy of the planet is: Select the correct answer choice: -GMám (1+1) -GMm (b) GMm (c) a+b (d) a b GMm(a-b) a+b²
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- (a) Imagine that a space probe could be fired as a projectile from the Earth's surface with an initial speed of 5.96 x 10“ m/s relative to the Sun. What would its speed be when it is very far from the Earth (in m/s)? Ignore atmospheric friction, the effects of other planets, and the rotation of the Earth. (Consider the mass of the Sun in your calculations.) 354790 Your response differs from the correct answer by more than 100%. m/s (b) What If? The speed provided in part (a) is very difficult to achieve technologically. Often, Jupiter is used as a "gravitational slingshot" to increase the speed of a probe to the escape speed from the solar system, which is 1.85 x 10“ m/s from a point on Jupiter's orbit around the Sun (if Jupiter is not nearby). If the probe is launched from the Earth's surface at a speed of 4.10 × 10“ m/s relative to the Sun, what is the increase in speed needed from the gravitational slingshot at Jupiter for the space probe to escape the solar system (in m/s)? (Assume…Consider the observation that the acceleration due to the gravitational force acting on a mass around a host planet decreases with the square of the separation between the objects. We can ask ourselves: why is it still accurate to consider a gravitational acceleration value of 9.8\frac{m}{s^2}9.8s2m for all of our projectile motion problems and all of our gravitational potential energy from prior modules? Let's analyze a situation and justify this analysis method: consider an object being launched from ground level to an altitude of 10,000 meters, roughly the cruising altitude of most jet liners, and far above our everyday experiences on Earth's surface. Compare the gravitational acceleration of the object at Earth's surface (the radius of Earth is about r_E=6.37\times10^6mrE=6.37×106m) to the acceleration value at the 10,000 meter altitude by determining the following ratio: g10,000m/gsurfaceA meteoroid is moving towards a planet. It has mass m = 0.54×109 kg and speed v1 = 4.7×107 m/s at distance R1 = 1.6×107 m from the center of the planet. The radius of the planet is R = 0.78×107 m. The mass of the planet is M = 5.6×1025kg. There is no air around the planet. a)Enter an expression for the total energy E of the meteoroid at R, the surface of the planet, in terms of defined quantities and v, the meteoroid’s speed when it reaches the planet’s surface. Select from the variables below to write your expression. Note that all variables may not be required.α, β, θ, d, g, G, h, m, M, P, R, R1, t, v, v1 b)Enter an expression for v, the meteoroid’s speed at the planet’s surface, in terms of G, M, v1, R1, and R. c)Calculate the value of v in meters per second.
- 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…At a particular instant, in a coordinate frame chosen by us, the Earth (mass = 6 x 1024 kg) is at location (-7.30 × 106, -7.80 × 106, 3.73 × 106) m and the Moon (mass = 7 x 10²² kg) is at location (-1.35 × 108, -2.50 × 10³, 2.82 × 108) m. A satellite with mass 5000 kg is at location (-1.18 × 10³, -2.73 × 10³, 3.10 × 108) m. What is the net force on the satellite? Assume that the gravitional force of the Sun is negligible. net = i i !)NA star with mass M and radius R collides head-on with another star of mass ¾*M and radius 4/5*R, and they coalesce to form a new start at rest whose radius is 6/5*R. Assume that initially the colliding stars had angular velocities with opposite directions but the same magnitude w. What is the magnitude and direction of the final’s stars angular velocity? (Express the magnitude as a fraction of w.)
- D Gm₁m₂ Fg KE = mv², Ug = - 2πr , ac = =²₁, v = ²7₁ T Gm₁m₂ GM g = G, Vesc = 2GM R , E = KE + Ug, G = 6.674 x 10-¹1 Nm²/kg² Problem 1: You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its radius to be 9 x 106 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 2.2 x 10¹¹ m from its star with a period of 402 days (3.473 x 107 sec). Once on the surface you find that the free-fall acceleration is 12.2 m/sec². a) What is the mass of the planet? Answer: 1.5 x 1025 kg. b) What is the mass of the star? Answer: 5.2 x 1030 kg.A meteoroid is moving towards a planet. It has mass m = 0.18×109 kg and speed v1 = 3.8×107 m/s at distance R1 = 1.6×107 m from the center of the planet. The radius of the planet is R = 0.26×107 m. The mass of the planet is M = 10×1025 kg. There is no air around the planet. a)Enter an expression for the total energy E of the meteoroid at R, the surface of the planet, in terms of defined quantities and v, the meteoroid’s speed when it reaches the planet’s surface. b)Enter an expression for v, the meteoroid’s speed at the planet’s surface, in terms of G, M, v1, R1, and R. c)Calculate the value of v in meters per second.A binary-star system contains a visible star and a black hole moving around their center of mass in circular orbits with radii r1 and r2 , respectively. The visible star has an orbital speed of v=5.36x105 ms-1 and a mass of m1 =5Ms ,where Ms= 1.98x1030kg is the mass of our Sun. Moreover, the orbital period of the visible star is T = 30 hours.(a) What is the radius r1 of the orbit of the visible star?(b) Calculate the mass m2 of the black hole in terms of MS . [Hint: One root of the equation x3 = 20a(5a+x)2 , where a is a constant, is x = 28a .]
- An object of mass m is released from rest a distance R above the surface of a planet of mass M and radius R. Derive an expression for the speed with which it hits the planet’s surface v.A certain asteroid in space can travel 48.0 km in 3.0 seconds. Its kinetic energy is known to be 3.84 x 1018 Joules. What is the mass of the asteroid?question 20 in the image