Answer:
1.129×10⁻⁵ N
1.295 m
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.
Answer:
C) one-half as great
Explanation:
We can calculate the acceleration of gravity in that planet, using the following kinematic equation:

In this case, the sphere starts from rest, so
. Replacing the given values and solving for g':

The acceleration due to gravity near Earth's surface is
. So, the acceleration due to gravity near the surface of the planet is approximately one-half of the acceleration due to gravity near Earth's surface.
Answer: V = 504m/a
F = 4N
Explanation: please find the attached file for the solution
Answer:
F=m(11.8m/s²)
For example, if m=10,000kg, F=118,000N.
Explanation:
There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

Solving for the force F, we obtain that:

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

In this case, the force from the rocket's thrusters is equal to 118,000N.