Question

PART ONE

Objects with masses of 207 kg and 516 kg
are separated by 0.482 m. A 31.8 kg mass is
placed midway between them.
Find the magnitude of the net gravitational
force exerted by the two larger masses on the
31.8 kg mass. The value of the universal gravitational constant is 6.672 × 10−11 N · m^2/kg^2.
Answer in units of N.

PART TWO

Leaving the distance between the 207 kg and
the 516 kg masses fixed, at what distance from
the 516 kg mass (other than infinitely remote ones) does the 31.8 kg mass experience a net force of zero?
Answer in units of m.

Answer 1

The magnitude of the net gravitational force between the masses is 3.38 N, and the distance from the 516 kg mass where the 31.8 kg mass experiences zero net force is 1.049 m.

Step-by-step explanation:

To find the magnitude of the net gravitational force exerted by the two larger masses on the 31.8 kg mass, we can use Newton's Law of Universal Gravitation:

F = G * ((m1 * m2) / r^2)

where F is the force, G is the universal gravitational constant (6.672 × 10⁻¹¹ N · m²/kg²), m1 and m2 are the masses, and r is the separation between the masses. Plugging in the values, the net gravitational force is:

F = 6.672 × 10⁻¹¹ * ((207 kg * 31.8 kg) / (0.241 m)^2) = 3.38 N

To find the distance from the 516 kg mass where the 31.8 kg mass experiences a net force of zero, we can use the same equation and solve for r:

0 = 6.672 × 10⁻¹¹ * ((516 kg * 31.8 kg) / (r)^2)

Simplifying, we get r = sqrt((516 kg * 31.8 kg) / (6.672 × 10⁻¹¹)) = 1.049 m

Answer 2

1.129×10⁻⁵ N

1.295 m

Step-by-step explanation:

Take right to be positive. Sum of forces on the 31.8 kg mass:

∑F = GM₁m / r₁² − GM₂m / r₂²

∑F = G (M₁ − M₂) m / r²

∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²

∑F = 1.129×10⁻⁵ N

Repeating the same steps, but this time ∑F = 0 and we're solving for r.

∑F = GM₁m / r₁² − GM₂m / r₂²

0 = GM₁m / r₁² − GM₂m / r₂²

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

516 / r² = 207 / (0.482 − r)²

516 (0.482 − r)² = 207 r²

516 (0.232 − 0.964 r + r²) = 207 r²

119.9 − 497.4 r + 516 r² = 207 r²

119.9 − 497.4 r + 309 r² = 0

r = 0.295 or 1.315

r can't be greater than 0.482, so r = 0.295 m.