What is geff, the effective free-fall acceleration, at a point on the earth's surface nearest the black hole? The earth itself accelerates, so geff is the acceleration of a mass at the surface relative to the earth's acceleration. Express your answer in meters per second squared.
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- ) Several planets possess nearly circular surrounding rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ringlike structures. Consider a homogeneous ring of mass M and radius R. a) What gravitational attraction does it exert on a particle of mass m located a distance x from the center of the ring along its axis? b) Suppose the particle falls from rest as a result of the attraction of the ring of matter. Find an expression for the speed with which it passes through the center of the ring. (a: see notes from class, b: Use the definition of potential energy.)Nothing can escape the event horizon of a black hole, not even light. You can think of the event horizon as being the distance from a black hole at which the escape speed is the speed of light, 3.00×10^8 m/s, making all escape impossible. What is the radius of the event horizon for a black hole with a mass 3.5 times the mass of the sun?Hunting a black hole. Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has moves in a circle of radius r1 and has orbital period T. Variations in the brightness of nearby stars suggest that the unseen companion moves in a circle of radius r2 (see the figure). Find the approximate masses (a) m1 of the visible star and (b) m2 of the dark star. Express your answer in terms of r1, r2, T, and G. I asked this question before and recieved an incorrect answer, so I'm asking again.
- 1. Consider a mass m initially at rest at a large distance I from center of the earth (l>R the earth's radius). The mass m is released and falls toward the earth. (a) Calculate the speed of the mass as a function of its distance x from the center of the earth. (b) In the approximation that l>>R, how much time does it take for the mass m to reach the earth? Express your answers in terms of R, 1, g (acceleration at the earth's surface) and (for part a) x.(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…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…
- Two planets P, and P, orbit around a star S in circular orbits with speeds v, = 46.4 km/s, and v, = 57.4 km/s respectivel (a) If the period of the first planet P, is 710 years what is the mass, in kg, of the star it orbits around? kg (b) Determine the orbital period, in years, of P2. yrLet's say we have a M1 and M2. Let's just say. Let's examine this hypothetical situation. The first mass would be 1.50 kg and the second mass would be 2.00 kg. These two masses would then be separated by a length or we should say a distance of this L = 2.50 m. Let's say we want to place a third mass (immaterial mass) in the middle of the two masses, such that there would be no net force. Find this specific place where this would occur, and to make it easy, find the distance from the first mass towards this third massBlack holes are difficult to observewith telescopes because they, bydefinition, don’t emit or reflect any light. They can be found by look-ing for other nearby objects orbit-ing them, however. Here is a dia-gram of a star in a circular orbit around a black hole. a. The period of the star’s orbit is 90 days, and its orbital radius around the black hole isobserved to be 3.6 : ×10^11 m. Find the orbital velocity of the star in units of m/s. (You need to convert 90 days to seconds, first). The circumference of a circle is 2πr. b. The mass of the star is known to be 4 × 10^30 kg. Find the centripetal acceleration of thestar and the strength of the gravitational force on the star. c. Find the mass of the black hole.
- Please solveMY NOTES ASK YOUR TEACHER PRACTICE ANOTHER (a) A spaceship is projected vertically upward from the Earth's surface with an initial speed of 6.91 km/s, but unfortunately does not have a great enough speed to escape Earth's gravity. What maximum height does the spaceship reach (in m)? Ignore air resistance. 3952535 (b) A meteoroid falls from a height of 1.91 x 107 m above the surface of the Earth. What is the speed (in m/s) when the meteorite hits the Earth? Assume the meteoroid is initially at rest with respect to the Earth. (Note that a meteorite is a meteoroid that makes it to Earth's surface.) 10698.65 X The fall of the meteorite is the time-reversal of the upward flight of the projectile, so it is described by the same energy equation as in part (a). m/s. Need Help? Read It SiThe planet, named 581 c that has a radius that is 1.5 times the radius of Earth. If the acceleration of the gravity on the surface of the planet 581 c, g 581c = 2.22 g Earth. Find the mass of the planet 581 c.