Question

SU Problem In 1994, the performer Rod Stewart drew over 3.0 million people to a concert in Rio de Janeiro, Brazil. The people in the group had an average mass of 80 kg. Standard E Problem 1) What collective gravitational force would the group have on a 5.5-kg eagle soaring 270 m above the throng? If you treat the group as a point object, you will get an upper limit for the gravitational force. (Express your answer to two significant figures.) Standard Exe Problem 1.2*10^(-6) UN Submit You currently have o submissions for this question. Only 10 submission are allowed. You can make 4 more submissions for this question. Standard Exer Problem Standard Exerc Problem 1 Your submissions: 1.210^(-6) Computed value: 1. 1*10^(-6) Xitted: Friday, November 22 at 11:34 PM Feedback: It appear 1.2 1096 X power of ten error. 1.2*10^(-6) X 310^-7 X 2) What is the ratio 3.3*10^- 7 X traction to the force between Earth and the eagle? (Express your answer to two significant figures.

Please help me. None of the above is the correct answer and I need urgent help. Thanks a lot!!!

Answer 1

Average mass of 1 person = 80 kg

No. of persons = 3.0 million = 10^6

mass of the total person, M = 80 x 10^6 kg

mass of the eagle, m = 5.5 kg

Distance between the group of people and the eagle, R = 270 m

Gravitational force between the group of people and the eagle,

F = G*M*m / R^2

where, G = Universal gravitational constant = 6.674 x 10^-11 m^2 / kg^2

Therefore,

F = (6.674 x 10^-11 x 80 x 10^6 x 5.5) / 270^2

= 4.028 x 10^-2 x 10^-5

= 0.4028 x 10^-6 N = 0.4028 $\mu$ N (Answer)