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Problems. Chapter 12. Mr. Newton, whose mass is 60.0 kg, is doing a physics demonstration in the front of the classroom. A) how much gravitational force does he exert on 55.0 kg Martha in the front row, 1.50 m away? F = G*m 1 *m 2 d 2. Variables/Given.
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Problems Chapter 12
Mr. Newton, whose mass is 60.0 kg, is doing a physics demonstration in the front of the classroom. • A) how much gravitational force does he exert on 55.0 kg Martha in the front row, 1.50 m away? • F = G*m1*m2 d2
Variables/Given • G = 6.67x10-11Nm2/kg2 • m1 = 60.0 kg • m2 = 55.0 kg • d = 1.50 m • F=?
Solution • F = (6.67x10-11Nm2/kg2)(60.0kg)(55.0kg) (1.50m)2 • F = 9.78x10-8 N
Mr. Newton, whose mass is 60.0 kg, is doing a physics demonstration in the front of the classroom. • B) how does this compare to what he exerts on 65.0 kg Lester, 4.00 m away in the back row?
Variables/Given • G = 6.67x10-11Nm2/kg2 • m1 = 60.0 kg • m2 = 65.0 kg • d = 4.00 m • F=?
Solution • F = (6.67x10-11Nm2/kg2)(60.0kg)(65.0kg) (4.00m)2 • F = 1.63x10-8 N
Our galaxy, the Milky Way, contains approximately 4.0x1011 stars with an average mass of 2.0x1030 kg each. How far away is the Milky Way from out nearest neighbor, the Andromeda Galaxy, if Andromeda has an average mass of 8x1041 kg and attracts the Milky Way with a gravitational force of 2.4x1030 N?
Variables/Given • G = 6.67x10-11Nm2/kg2 • m1 = (2.0x1030kg)(4.0x1011) = 8x1041 kg • m2 = 8x1041 kg • F = 2.4x1030 N • d=?
Solution • d = √G*m1*m2/F • d= √(6.67x10-11)(8x1041)(8x1041)/(2.4x1030) • d = 4.2x1021m
Gravitational Acceleration • Force= mass * acceleration (gravity) • m * g = G * m1 * m2 d2 • One of the masses cancels, left with • g = G * m2 d2
Teresa is standing in the lunch line 6.38x106m from the center of Earth. Earth’s mass is 5.98x1024kg. • What is the acceleration due to gravity?
Variables/Given • m2=5.98x1024 kg • d=6.38x106m • G=6.67x10-11N*m2/kg2 • g=?
Solution • g = G * m2 d2 • g = (6.67x10-11)(5.98x1024) (6.38x106)2 • g = 9.80 m/s2