Objects with masses of 280 kg and a 580 kg are separated by 0.340 m. (a) Find th
ID: 1627902 • Letter: O
Question
Objects with masses of 280 kg and a 580 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objects on a 33.0-kg object placed midway between them. magnitude N direction (b) At what position (other than infinitely remote ones) can the 33.0-kg object be placed so as to experience a net force of zero? m from the 580-kg mass Two objects attract each other with a gravitational force of magnitude 9.50 times 10^-9 N when separated by 19.3 cm. If the total mass of the objects is 5.10 kg, what is the mass of each? heavier mass Your response differs from the correct answer by more than 10%. Double check your calculations. kg lighter mass Your response differs from the correct answer by more than 100%. kg A synchronous satellite, which always remains above the same point on a planet's equation, is put in circular orbit around Mars so that scientists can study a surface feature. Mars rotates once every 24-6 h, use the data of this table to find the altitude of the satellite. km.Explanation / Answer
Let the 2 original masses be m1, m2 separated by distance d, and the third mass be m3 at distance x from m2, and G be the gravitational constant.
The resultant force F towards m2 is:
F = G m2 m3 / x^2 - G m1 m3 / (d - x)^2
G m3 [ m2 / x^2 - m1 / (d - x)^2 ] ...(1)
(a) Putting x = d / 2 in (1):
F = 4G m3 (m2 - m1) / d^2
F = 4 * 6.673*10^(- 11) * 33*(580-280) / 0.340^2
F = 2.286*10^-5 N
(b) Equating F to 0 in (1):
m2 / x^2 = m1 / (d - x)^2
m1 x^2 = m2(d^2 + x^2 - 2dx)
(m2 - m1)x^2 - 2d m2 x + m2 d^2 = 0
x = (2d m2 +/- sqrt[ 4d^2 m2^2 - 4m2 d^2 (m2 - m1) ]) / [ 2(m2 - m1)
x = d( m2 +/- sqrt[ m1 m2 ]) / (m2 - m1)
The negative sign gives the root between 0 and 0.340:
x = 0.340*(580 - sqrt(280 * 580))/ (580 - 280)
x = 0.2 m.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.