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 × 10%, -2.50 × 10³, 2.82 × 108) m. A satellite with mass 5000 kg is at location (-1.18 x 108, -2.73 × 108, 3.10 × 108) m. What is the net force on the satellite? Assume that the gravitional force of the Sun is negligible. net = ! i >N
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- During a solar eclipse, the Moon is positioned directly between Earth and the Sun. The masses of the Sun, Earth, and the Moon are 1.99×10^30 kg, 5.98×10^24 kg, and 7.36×10^22 kg, respectively. The Moon's mean distance from Earth is 3.84×10^8 m, and Earth's mean distance from the Sun is 1.50×10^11 m. The gravitational constant is G=6.67×10^−11 N·m2/kg^2. Find the magnitude F of the net gravitational force acting on the Moon during the solar eclipse due to both Earth and the Sun.Astronomical observations of our Milky Way galaxy indicate that it has a mass of about 8.0 x 1011 solar masses. A star orbiting near the galaxy's periphery is 5.6 x 104 light-years from its center. (a) What should the orbital period (in y) of that star be? y (b) If its period is 5.3 x 107 years instead, what is the mass (in solar masses) of the galaxy? Such calculations are used to imply the existence of other matter, such as a very massive black hole at the center of the Milky Way. solar massesA planet of mass 5 ⨯ 1024 kg is at location <4 ⨯ 1011, −4 ⨯ 1011, 0> m. A star of mass 4 ⨯ 1030 kg is at location <−6 ⨯ 1011, 4 ⨯ 1011, 0> m. (a) What is the relative position vector pointing from the planet to the star? (b) What is the distance between the planet and the star? (c) What is the unit vector in the direction of r? (d) What is the magnitude of the force exerted on the planet by the star?(e) What is the magnitude of the force exerted on the star by the planet? (f) What is the force (vector) exerted on the planet by the star? (g) What is the force (vector) exerted on the star by the planet? (Note the change in units.)
- An astronomer with a mass of (1.7x10^2) kg drifts in deep space along another astronomer with mass of (1.00x10^2) kg. The distance between the two is (9.000x10^0) m. What is the gravitational force between the two astronomers?Two spherical objects have a combined mass of 160 kg . The gravitational attraction between them is 7.61×10−6 NN when their centers are 21.0 cm apart. What is the mass of the heavier object? What is the mass of the lighter object?If you weigh 685 N on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 18.0 km ? Take the mass of the sun to be ms = 1.99×1030 kg , the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s2 . Express your weight wstar in newtons.
- John (70.0 kg) is standing 2.50 m away from Sam (60.0 kg). Find the gravitational force between John and Sam. Possible Formulas that can be used to solve the question: v=(2πr)/T ac=v2/r ac=(4π2r)/T2 Fc=mac Fg=mg F=(Gm1m2)/d2 g=Gm/r2 T2=(4π2/Gm)r3 v=√(Gm)/r g=9.80m/s2 G=6.67x10-11 (N∙m2)/kg2During a solar eclipse, the Moon is positioned directly between Earth and the Sun. The masses of the Sun, Earth, and the Moon are 1.99 x 1030 kg, 5.98 × 1024 kg, and 7.36 x 1022 kg, respectively. The Moon's mean distance from Earth is 3.84 × 10 m, and Earth's mean distance from the Sun is 1.50 × 10'" is G = 6.67 x 10-1' N•m²/kg². m. The gravitational constant Find the magnitude F of the net gravitational force acting on the Moon during the solar eclipse due to both Earth and the Sun. 2.6 x1045 F = N IncorrectWhen the Moon is directly overhead at sunset, the force by Earth on the Moon, FEM, is essentially at 90° to the force by the Sun on the Moon, FSM, as shown below. FEM=1.98×10^20N and FSM=4.36×10^20N, all other forces on the Moon are negligible, and the mass of the Moon is 7.35×10^22kg. 1. Write an expression for the magnitude of the acceleration of the Moon. 2. What is the acceleration of the Moon in m/s^2?
- 1Plaskett'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…Assume the data collected were as follows: (where the force in Newtons is computed by converting the mass to kilograms and then multiplying by the gravitational acceleration 9.8 m/s²) m (grams) Force (Newtons) x (cm) (measured) (computed) (measured) 1000 9.8 1 2000 19.6 1.5 3000 29.4 2.1 4000 39.2 2.4 5000 49 3 Make a plot of force vs. stretch distance x. This means force should be on the vertical axis. Since Fgravity = -Fspring (they are in opposite directions) and Fspring = -kx, then: Fgravity = kx This means the slope of your graph is the spring constant k. Fill in your value for k (measured in Newtons/cm)