Question

1. (2 points) As shown in the figure to the right, a 2 kg mass is placed at x 3 cm, y-5 cm. A 7 kg mass is placed at x 9 cm, y 2 cm. A 5 kg mass is placed at the origin. What is the magnitude and direction of the net gravitational force on the 2 kg mass from the 5 kg and 7 kg masses? (Assume that these masses are located in space, i.e. ignore the weight force from Earth). 10 2 kg 7 kg 2. Is it possible to place a satellite in orbit so that it is always directly above 60 N latitude? Explain why or why not 1 2 3 4 5 6 7 8 9 10

Answer 1

Given that As shown the figure is right Now find r_12 & r_23 therefore r_12 = squareroot 5^2 + 3^2 = squareroot 25 + 9 \r\n = squareroot 345 = 5.83 cm theta_1 = Tan^-1 (3/5) = 30.96 degree r_23 = squareroot 3^2 + 6^2 = squareroot 9 + 36 = squareroot = 45 = 6.71 cm theta_2 = Tan^-1 (6/3) = 63.43 degree Now, gravitational force on m_2 due to m_1 is f_12 = G m_1 m_2/r_12^2 = 6.67 times 10^-11 times 5 times 2/(5.83 times 10^-2)^2 \r\n = 1.962 times 10^-7 kg m/s^2 gravitational force on m_2 due to m_3 f_23 = G m_2 m_3/r_2 3^2 = 6.67 times 10^-11 times 7 times 2/(6.71 times 10^-2)^2 = 2.074 times 10^-7 kg m/s^2 theta = theta_1 + theta_2 = 94.395^mid dot therefore The resultant force f on mass m_2 is given by f = squareroot f_12^2 + f_23^2 + 2 f_12 f_23 cos theta = squareroot 1.962^2 + 2.074^2 + 2