/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 55 An astronaut, whose mission is t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

An astronaut, whose mission is to go where no one has gone before, lands on a spherical planet in a distant galaxy. As she stands on the surface of the planet, she releases a small rock from rest and finds that it takes the rock 0.480 s to fall 1.90 m. If the radius of the planet is 8.60 \(\times\) 10\(^7\) m, what is the mass of the planet?

Short Answer

Expert verified
The mass of the planet is approximately \(4.87 \times 10^{24}\,\text{kg}\).

Step by step solution

01

Determine the acceleration due to gravity

To find the acceleration due to gravity (\(g\)) on the planet, we use the formula for the distance covered by an object under constant acceleration: \[d = \frac{1}{2} g t^2\]Here, \(d = 1.90\,\text{m}\) and \(t = 0.480\,\text{s}\). Plugging in these values, we get:\[1.90 = \frac{1}{2}g (0.480)^2\]Solving for \(g\), we find:\[g = \frac{2 imes 1.90}{0.480^2}\approx 16.46\,\text{m/s}^2\]
02

Use the formula for gravitational force

The gravitational force acting on a body is given by the equation:\[F = \frac{G M m}{r^2}\]where \(G = 6.674 \times 10^{-11}\,\text{Nm}^2/\text{kg}^2\) is the gravitational constant, \(M\) is the mass of the planet, \(m\) is the mass of the rock (which cancels out later), and \(r\) is the radius of the planet.
03

Relate acceleration due to gravity to gravitational force

The acceleration due to gravity \(g\) can be expressed as:\[g = \frac{G M}{r^2}\]We know \(g = 16.46\,\text{m/s}^2\) from Step 1 and \(r = 8.60 \times 10^7\,\text{m}\). Solving for \(M\), the mass of the planet, we equate:\[16.46 = \frac{6.674 \times 10^{-11} M}{(8.60 \times 10^7)^2}\]
04

Solve for the mass of the planet

Rearrange the equation from Step 3 to solve for \(M\):\[M = \frac{16.46 \times (8.60 \times 10^7)^2}{6.674 \times 10^{-11}}\]Calculate \(M\):\[M \approx 4.87 \times 10^{24}\,\text{kg}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acceleration Due to Gravity
Understanding acceleration due to gravity is crucial in comprehending how objects behave under the influence of gravity. On different planets, this acceleration can vary due to the mass and radius of the planet. This force is the key reason why objects fall towards the surface.
In physics, the acceleration due to gravity is often symbolized by the letter \( g \). On Earth, we experience an average \( g \) of approximately 9.81 m/s². However, this value changes on other celestial bodies. For instance, a smaller or less massive planet will exert less gravitational pull, resulting in a lower \( g \). Conversely, a more massive planet will have a higher \( g \). Understanding this variation is essential when calculating movements and trajectories on different planetary surfaces.
In the given exercise, using the formula \(d = \frac{1}{2} g t^2\), where \(d\) is distance and \(t\) is the time, helped us determine the acceleration due to gravity on the new planet. By rearranging and solving this equation, we found that \( g \) for this planet is 16.46 m/s², which is significantly higher than on Earth.
Spherical Planets
When considering celestial bodies like planets, their shape plays a role in how we calculate gravitational forces. Most planets, including the one in our exercise, are approximately spherical in shape. This assumption simplifies calculations because the distance from the planet’s center to any point on its surface (the radius) is consistent.
Knowing the radius is pivotal since it affects the calculation of gravitational force. Therefore, the formula for gravitational force includes the radius: \( F = \frac{G M m}{r^2} \), where \( r \) is the planet's radius.
In our exercise, with the given radius of 8.60 × 10 extsuperscript{7} m, we combine this with our previously calculated \( g \) to solve for the planet's mass. Due to being spherical, we apply standard spherical formulas without needing complex geometry adjustments, making calculations far more manageable.
Newton's Laws of Motion
Newton's Laws of Motion, especially his universal law of gravitation, help us understand the movement of objects and forces acting within our universe. Newton's law states that every point mass attracts every other point mass with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This principle forms the foundation of calculating gravitational force.
The formula \( F = \frac{G M m}{r^2} \) arises from this universal law. In our exercise, the astronaut determines gravitational acceleration which is crucial for deriving the planet’s mass.
Newton’s formulations allow predicting how quickly the rock falls and the net forces acting upon it. In this respect, laws enabling the understanding of free fall and gravity provide tools for broader applications, from space travel to engineering on Earth. Newton’s contribution makes it possible to calculate variables like the mass of a planet by merely observing an object's motion under gravity.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Some scientists are eager to send a remote-controlled submarine to Jupiter's moon Europa to search for life in its oceans below an icy crust. Europa's mass has been measured to be 4.80 \(\times\) 10\(^{22}\) kg, its diameter is 3120 km, and it has no appreciable atmosphere. Assume that the layer of ice at the surface is not thick enough to exert substantial force on the water. If the windows of the submarine you are designing each have an area of 625 cm\(^2\) and can stand a maximum inward force of 8750 N per window, what is the greatest depth to which this submarine can safely dive?

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at 30,000 km/s. (a) How far are these clumps from the center of the black hole? (b) What is the mass of this black hole, assuming circular orbits? Express your answer in kilograms and as a multiple of our sun's mass. (c) What is the radius of its event horizon?

The star Rho\(^1\) Cancri is 57 light-years from the earth and has a mass 0.85 times that of our sun. A planet has been detected in a circular orbit around Rho\(^1\) Cancri with an orbital radius equal to 0.11 times the radius of the earth's orbit around the sun. What are (a) the orbital speed and (b) the orbital period of the planet of Rho\(^1\) Cancri?

A uniform, solid, 1000.0-kg sphere has a radius of 5.00 m. (a) Find the gravitational force this sphere exerts on a 2.00-kg point mass placed at the following distances from the center of the sphere: (i) 5.01 m, (ii) 2.50 m. (b) Sketch a qualitative graph of the magnitude of the gravitational force this sphere exerts on a point mass \(m\) as a function of the distance \(r\) of \(m\) from the center of the sphere. Include the region from \(r = 0\) to \(r\) \(\rightarrow\) \(\infty\).

In March 2006, two small satellites were discovered orbiting Pluto, one at a distance of 48,000 km and the other at 64,000 km. Pluto already was known to have a large satellite Charon, orbiting at 19,600 km with an orbital period of 6.39 days. Assuming that the satellites do not affect each other, find the orbital periods of the two small satellites \(without\) using the mass of Pluto.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.