A spaceship with a mass of $2.8 \times 10 ^ { 6 } \mathrm {~kg}$ mass of $5.0 \times 10 ^ { 16 } \mathrm {~kg}$ joining the asteroids and is aimed at the midpoint of that line. What is the net gravitational force exerted by the asteroids on the spaceship when the spaceship is 30 \mathrm{~km} away from that midpoint?
$\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } \right)$
A) 16,000 N
B) 6,200 N
C) 18,000 N
D) 8,000 N
E) 12,000 N

Question

A spaceship with a mass of $2.8 \times 10 ^ { 6 } \mathrm {~kg}$ mass of $5.0 \times 10 ^ { 16 } \mathrm {~kg}$ joining the asteroids and is aimed at the midpoint of that line. What is the net gravitational force exerted by the asteroids on the spaceship when the spaceship is  30 \mathrm{~km}  away from that midpoint?
$\left( G = 6.67 \times 10 ^ { - 11 } \mathrm {~N} \cdot \mathrm { m } ^ { 2 } / \mathrm { kg } ^ { 2 } \right)$
A) 16,000 N
B) 6,200 N
C) 18,000 N
D) 8,000 N
E) 12,000 N

Quizplus · Accepted Answer

The gravitational force $F$ between two masses $m_1$ and $m_2$ separated by a distance $r$ is given by Newton's law of universal gravitation: $F = G \frac{m_1 m_2}{r^2}$, where $G$ is the gravitational constant. In this scenario, the spaceship is attracted to two asteroids of equal mass $5.0 \times 10^{16} \mathrm{~kg}$ each. The spaceship is 30 km (or 30,000 m) away from the midpoint between the two asteroids, meaning it is 30,000 m from each asteroid along the line connecting them.The gravitational force exerted by each asteroid on the spaceship can be calculated using the formula given. Since the spaceship is equidistant from both asteroids, the forces exerted by each asteroid will be equal in magnitude but will act along the line connecting the spaceship to each asteroid, towards each asteroid.Calculating the force due to one asteroid (since the situation is symmetric, the total force will be twice the force calculated for one asteroid):$$F = G \frac{m_{spaceship} m_{asteroid}}{r^2}$$$$F = 6.67 \times 10^{-11} \frac{(2.8 \times 10^6)(5.0 \times 10^{16})}{(30,000)^2}$$$$F = 6.67 \times 10^{-11} \frac{(2.8 \times 10^6)(5.0 \times 10^{16})}{900,000,000}$$$$F = 6.67 \times 10^{-11} \times \frac{1.4 \times 10^{23}}{9 \times 10^8}$$$$F = 6.67 \times 10^{-11} \times 1.55555556 \times 10^{14}$$$$F \approx 1.037 \times 10^4 \, \text{N}$$Since the total force is the sum of the forces exerted by each asteroid, and both forces are equal and additive in this configuration, the total force is:$$F_{total} = 2 \times F \approx 2 \times 1.037 \times 10^4 \, \text{N} \approx 2.074 \times 10^4 \, \text{N}$$Rounding to the nearest thousand gives approximately 20,000 N, which is not among the provided options. It seems there was a mistake in my calculation or interpretation of the options. Given the setup and the calculation approach, the correct answer should logically follow from the gravitational force equation, but none of the provided options exactly match the expected result based on the calculation. Re-evaluating the calculation steps or the interpretation of the question might be necessary.

A Spaceship with a Mass Of 2.8×106 kg2.8 \times 10 ^ { 6 } \mathrm {~kg}2.8×106 kg

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