Objects with masses of 181 kg and 712 kg are separated by 0.442 m. A 72.6 kg mass is placed midway between them. 0.442 m b b 181 kg 72.6 kg 712 kg Find the magnitude of the net gravitational force exerted by the two larger masses on the 72.6 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m2 /kg2 . Answer in units of N

Respuesta :

Answer:

The Force exerted by the two objects will be same and 4,401,189.49  × 10−11 N

Explanation:

Let m1 be the mass of the first object and m2 be the mass of the second object and the distance between them be d. Then

m1= 181kg

m2= 712 kg

d= 0.442 m

G is the gravitational Constant

According to Newtons Law Gravitational Force is given by

F=G m1 m2 /d²

F= 6.672 × 10−11 ×181×712/(0.442)²

F=  859,576.24  × 10−11/0.195364

F=  4,401,189.49  × 10−11 N

The Force exerted by the two masses on the third mass will be same and 4,401,189.49  × 10−11 N

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