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Leaving the distance between the 181 kg and the 712 kg masses fixed, at what distance from the 712 kg mass (other than infinitely remote ones) does the 72.6 kg mass experience a net force of zero? Answer in units of m

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Final answer:

This Physics problem requires the use of the universal law of gravitation to determine the exact distance at which a 72.6 kg mass experiences no net gravitational force from two other fixed masses.

Step-by-step explanation:

The question concerns the concept of gravitational force and specifically asks at what distance a 72.6 kg mass should be placed from a 712 kg mass so that it experiences a net gravitational force of zero, given that the distance between the 181 kg and 712 kg masses is fixed. This is a problem that entails using the universal law of gravitation, which states that the force between two masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

To solve this problem, the net force on the 72.6 kg mass due to the two other masses (181 kg and 712 kg) must be set to zero. By establishing two equations that represent the gravitational forces exerted on the 72.6 kg mass by the other two masses and setting them equal to one another, we can solve for the required distance. Therefore, we'll use the gravitation equation F = G(m1*m2)/r², where G is the gravitational constant, m1 and m2 are the masses, and r is the distance between the centers of the masses.

User Walt Reed
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Answer:

Step-by-step explanation:

The force due to gravitation is equal to zero for each of the masses.

M1= 181kg

M2= 712kg

m = 72.6kg

The distance between M1 and M2 is said to be fixed , therefore no value should be given I.e it's a constant.

From the formula for gravitational force we have that

F = GMm/r^2

GmM1/(d-r)^2. = GmM2/r^2

where r is the distance between the 72.6 kg and 712kg

d is the distance between M1 and M2

Solving mathematically

r(√M1+√M2) = d√M2

r = d√M2/√M1 + √M2

d×26.68/ 13.45+26.68

d×26.68/40.13

r = 0.665d

User Dickfala
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