Question

After a spacewalk, a \(1.00-\mathrm{kg}\) tool is left \(50.0 \mathrm{~m}\) from the center of gravity of a 20.0 -metric ton space station, orbiting along with it. How much closer to the space station will the tool drift in an hour due to the gravitational attraction of the space station?

Short answer

Answer: The tool drifts approximately 0.3472 meters closer to the space station in an hour due to gravitational attraction.

Step-by-step solution

Find the gravitational force

We use Newton's law of universal gravitation to find the gravitational force between the tool and the space station: \(F = G \frac{m_1 m_2}{r^2}\) Where \(F\) is the gravitational force, \(G\) is the universal gravitational constant (\(G \approx 6.674 \times 10^{-11} \mathrm{Nm^2/kg^2}\)), \(m_1\) and \(m_2\) are the masses of the objects, and \(r\) is the distance between them. First, we need to convert the mass of the space station to kilograms: \(20.0 \;\mathrm{metric\;tons} = 20,000 \;\mathrm{kg}\) Now we can plug in the values: \(F = (6.674 \times 10^{-11} \;\mathrm{Nm^2/kg^2}) \frac{(1.00 \;\mathrm{kg})(20,000 \;\mathrm{kg})}{(50.0 \;\mathrm{m})^2}\)

Calculate the gravitational force

Calculate the gravitational force: \(F = 5.3392 \times 10^{-8} \;\mathrm{N}\)

Determine the tool's acceleration

According to Newton's second law of motion, the force on an object equals the mass of the object, multiplied by the acceleration (\(F = ma\)). Solving for acceleration (a): \(a = \frac{F}{m}\) We can plug in values for the mass of the tool (1 kg) and the gravitational force: \(a = \frac{5.3392 \times 10^{-8} \;\mathrm{N}}{1.00 \;\mathrm{kg}}\)

Calculate the acceleration

Calculate the tool's acceleration due to the gravitational force: \(a = 5.3392 \times 10^{-8} \;\mathrm{m/s^2}\)

Compute the distance the tool drifts

Using the acceleration and the time (1 hour), we can compute the distance that the tool drifts towards the space station. We use the constant acceleration formula: \(d = v_0 t + \frac{1}{2}at^2\) Since the tool's initial velocity (\(v_0\)) relative to the space station is 0, the formula becomes: \(d = \frac{1}{2}at^2\) We need to convert the time to seconds: \(1 \;\mathrm{hour} = 3600 \;\mathrm{s}\) Now, plug the values for acceleration and time into the equation: \(d = \frac{1}{2}(5.3392 \times 10^{-8} \;\mathrm{m/s^2})(3600 \;\mathrm{s})^2\)

Calculate the distance drifted

Calculate the distance that the tool drifts towards the space station in an hour: \(d = 0.3472 \;\mathrm{m}\) The tool drifts approximately 0.3472 meters closer to the space station in an hour due to gravitational attraction.