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The planet Saturn has about 100 times the mass of the earth and is about 10 times farther from the sun than the earth is. Compared to the acceleration of the earth caused by the sun’s gravitational pull, how great is the acceleration of Saturn due to the sun’s gravitation? (i) 100 times greater; (ii) 10 times greater; (iii) the same; (iv)
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- Hi please help: Three identical very dense masses of 7600 kg each are placed on the x axis. One mass is x1= -140cm, one is at the origin, and one is at x2=450cm What is the magnitude of the net gravitational force Fgrav on the mass at the origin due to the other two masses? Take the gravitational constant to be G=6.67 x 10^-11N•m^2/kg^2 express answer in Newton's to three significant figures.arrow_forwardYou may have an image of Sir Isaac Newton sitting under a tree and after being hit on the head by an apple he suddenly "discovered" the Law of Universal Gravitation. In fact, the theory was a result of years’ worth of research, which in turn was based on centuries of accumulated knowledge. He is credited with determining that the following relationship is universal. The gravitational attraction between two objects varies jointly with their masses (m1 and m2) and inversely with the square of the distance (d) between them. By what percent does the force of gravitational attraction change if one mass is increased by 20%, the other mass decreased by 20%, and the separation is reduced by 25%?arrow_forwardSo let's consider a person with a mass of 51.0 kg standing on the Earth. To find the gravitational force on the person, we'll again use Newton's law of universal gravitation with the Earth as m2 and the radius of the Earth for the distance F = GmME RE2 . Now all we need to do is substitute values and calculate. We already said m = 51.0 kg, and we know G = 6.67 ✕ 10−11 N · m2/kg2. The Earth is not a perfect sphere, but, its average radius is RE = 6.37 ✕ 106 m. The mass of the Earth is ME = 5.97 ✕ 10−24 kg. We can then substitute these values in the following formula. (Enter your answer in N.) F = (6.67 ✕ 10−11 N · m2/kg2)(51.0 kg)(5.97 ✕ 1024 kg) (6.37 ✕ 106 m)2 (A) = _______ N Now let's compare this result to the person's weight (in N) found by multiplying the person's mass by g (or, that is, w = mg) where g = 9.80 m/s2. w = (51.0 kg)(9.80 m/s2) =(b) __________________ N You should have found that these two methods give about the same result!…arrow_forward
- A research team has discovered that a moon is circling a planet of our solar system: The moonorbits the planet once every 7 hours on a nearly circular orbit in a distance R of 48000 km fromthe centre of the planet. Unfortunately, the mass m of the moon is not known. Use Newton’s lawof gravitation with G = 6.67 · 10−11 m3/(kg·s2) to approach the following questions:F = G ·mMR2(1)(a) Based on the observations, determine the total mass M of the planet.(b) Which moon and planet of our solar system is the team observing? (Use literature.)arrow_forwardWonder Woman and Superman fly to an altitude of 1530 km, carrying between them a chest full of jewels that they intend to put into orbit around Earth. They want to make this tempting treasure inaccessible to their evil enemies who are trying to gain possession of it, yet keep it available for themselves for future use when they retire and settle down. But perhaps the time to retire is now! They accidentally drop the chest, which leaves their weary hands at rest, and discover that they are no longer capable of catching it as it falls into the Pacific Ocean. At what speed vf does the chest impact the surface of the water? Ignore air resistance (in reality, it would make large difference). The radius and mass of Earth are 6370 km and 5.98 x 1024 kg, respectively. m/s %3Darrow_forwardAn object of mass m is launched from a planet of mass M and radius R. a)Derive and enter an expression for the minimum launch speed needed for the object to escape gravity, i.e. to be able to just reach r = ∞. b)Calculate this minimum launch speed (called the escape speed), in meters per second, for a planet of mass M = 6 × 1023 kg and R = 76 × 104 km.arrow_forward
- Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0 • 1011 solar masses. A star orbiting on the galaxy’s periphery is about 6.0 • 104 light years from its center. a) What should the orbital period of that star be in years? b) If its period is 6.0 • 107 years instead, what is the mass of the galaxy in solar masses? Such calculations are used to imply the existence of “dark matter” in the universe and have indicated, for example, the existence of very massive black holes at the centers of some galaxies.arrow_forwardThe law of universal gravitation says that there is a force between two objects in the universe. Can there ever be a physical instance when the two objects will never have an influence on each other? Hint: look at the equation that relates force, mass, distance, and the gravitational constant G: Gxm, xma F = O No, no matter how far apart they are, there will always be some tiny force acting between the two. O No, Since the distance between two objects can never be zero. All listed answers here are correct. O No, they will always influence each other. You cannot divide by zero.arrow_forwardQuestion 4 of 7 GMm where If the gravitational force between two objects of mass M and m, separated by a distancer, has magnitude G = 6.67 × 10-1" m°kg¬'s¯², then the work required to increase the separation from a distance r¡ to a distance r2 is GMm(r,' – r,'). Compute the work required to move a 1500-kg satellite from an orbit 1000 km above the surface of Earth to an orbit 1500 km above the surface of Earth. Assume that Earth is a sphere of radius R = 6.37 × 10° m and mass M. = 5.98 × 1024 kg. Treat the satellite as a point mass. (Write your answer in scientific notation with two decimal places.) 1.47 x107 J W = Incorrect Question Source: Rogawski 4e Calculus Early Transcendentals| Publisher: W.H. Fm 014 étv hulu MacBook PrOarrow_forward
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