Three uniform spheres with masses m_1 = 6.0 kg, m_2 = 2.0 kg and m_1 = 4.0 kg ar

Question

Three uniform spheres with masses m_1 = 6.0 kg, m_2 = 2.0 kg and m_1 = 4.0 kg are fixed at the positions shown in the diagram. Assume they are completely isolated and there are no other masses nearby (a) What is the magnitude of the acceleration on a particle placed at the origin (Point P)? What is the direction of this acceleration? (b) For a particle at point P what speed is required for it to completely escape the gravitational attraction of the three fixed masses? (c) How much energy is required to separate the three masses so that they are very far apart? The exoplanet Poltergeist is approximately 2300 light years away and has a mass of about 2.6 times 10^25 kg. Assuming it has the same average density as Earth, what a its (a) surface gravity (the acceleration of gravity at its surface), and (b) escape speed from the surface? If you know the mass of a star and an object's speed at perihelion (closest approach to the star) and speed at aphelion (farthest point from the star), you can find the perihelion and aphelion distances. A particular comet has perihelion speed 8.3 times 10^4 m/s and aphelion speed 6.8 times 10^3 m/s as it orbits the Sun. (a) What are the closest and farthest distances from the star for this comet? What is the eccentricity of the orbit? (b) What is the expected orbital period of this comet? A planet revolves around a star in an elliptical orbit. At perihelion it has distance 7.8 times 10^10 m and speed 5.7 times 10^4 m/s and at aphelion it has distance 7.8 times 10^11 m and speed 5.7 times 10^3 m/s (a) What is the mass of the star? (b) What is the planet's speed when it is a distance of 2.5 times 10^11 m from the star? What is the angle between the radial vector and the velocity vector at this point? Sketch the orbit and indicate the approximate position and radial and velocity vectors at this point.

Explanation / Answer

10.1:

(A) field strength at P due to m1:

g1 = G m1 / r1^2 = (6.67 x 10^-11 ) (6) / (0.03^2)

g1 = 4.45 x 10^-7 m/s^2 along +y axis

due to m2 :

g2 = G m2 / r2^2 = (6.67 x 10^-11)(2) / (0.01^2)

= 1.334 x 10^-6 m/s^2 along -x axis.

due to m3:

g3 = (6.67 x 10^-11) (4) / (0.02^2) =6.67 x 10^-11 along -y axis

g_net = - 13.34 x 10^-7 i - 1.50 x 10^-7 j

magnitude of net acceleration = sqrt(13.34^2 + 1.50^2) x 10^-7

= 13.4 x 10^-7 m/s^2 .......Ans

(B) Applying energy conservation,

PE + KE = constant

- G m1 m / r1 - Gm2 m / r2 - G m3 m / r3 + m v^2 /2 = 0 + 0

v^2 /2 = (6.67 x 10^-11) [ 6/0.03 + 2/0.01 + 4 / 0.02]

v = 2.83 x 10^-4 m/s

(C) POtential energy of system,

PE = - (6.67 x 10^-11) [ 6 x 4 / 0.05 + 2 x 4 / sqrt(0.02^2 + 0.01^2) + 6 x 2 / sqrt(0.03^2 + 0.01^2)]

PE = - 7.81 x 10^-8 J

hence energy required = 7.81 x 10^-8 J

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