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A satellite of Mars, called Phoebus, has an orbital radius of \(9.4 \times 10^{6} \mathrm{~m}\) and a period of \(2.8 \times 10^{4} \mathrm{~s}\). Assuming the orbit is circular, determine the mass of Mars.

Short Answer

Expert verified
To find the mass of Mars, first equate the gravitational force to the centripetal force. Then substitute for velocity using the given orbital radius and period. Finally, calculate the mass by inserting the given values into the equation obtained. Make sure to use the correct units and unit conversions where necessary to get the exact mass of Mars.

Step by step solution

01

Identify Relevant Formulas

The orbital motion of an object (in this case, the satellite Phoebus) is ruled by the balance between the gravitational pull of the central object (Mars) and the required centripetal force for the orbiting motion. Accordingly, we utilize the following two formulas: Newton's law of Universal Gravitation given by \(F = G * (M * m) / r^2\) and the Centripetal Force formula represented by \(F = m * v^2 / r\). Here, F is the gravitational force, G is the gravitational constant, M and m are the masses of Mars and Phoebus respectively, r is the distance between Mars and Phoebus and v is the velocity of Phoebus.
02

Equate Forces and Rearrange for Mass

Since the gravitational force is providing the necessary centripetal force to the satellite, we can equate two forces and solve for Mars's mass (M). Thus, we obtain \(G * (M * m) / r^2 = m * v^2 / r\). Simplifying, we get \(M = v^2 * r / G\).
03

Substitute for the Velocity

The speed (v) of the satellite can be related to its orbital radius (r) and orbital period (T) by the relationship \(v = 2 * pi * r / T\). Substituting this into previous equation, Mars's mass (M) can be rewritten as \(M = (4 * pi^2 * r^3) / (G * T^2)\).
04

Substitute Given Values

To find the mass of Mars, insert the given values into the equation (G = \(6.674 * 10^{-11}\) N(m/Kg)², r = \(9.4 * 10^{6}\) m, and T = \(2.8 * 10^{4}\) s), and calculate to find \(M = (4 * pi^2 * (9.4 * 10^{6})^3) / ((6.674 * 10^{-11}) * (2.8 * 10^{4})^2)\).
05

Calculate the Result

Calculating the above expression will provide the mass of Mars.

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Key Concepts

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

Newton's Law of Universal Gravitation
Sir Isaac Newton's law of Universal Gravitation is a fundamental principle that describes the attractive force between two objects. According to this law, every point mass attracts every other point mass by a force acting along the line intersecting both points. This force is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.

The formula for Newton's law of Universal Gravitation is given by: \[F = G \frac{Mm}{r^2}\] where,
  • \(F\) is the gravitational force between the masses,
  • \(G\) is the gravitational constant \(6.674 \times 10^{-11} \mathrm{Nm}^2/\mathrm{kg}^2\),
  • \(M\) and \(m\) are the masses of the two objects,
  • \(r\) is the distance between the centers of the two masses.
Understanding this concept is crucial to gravitation and celestial mechanics as it explains why planets orbit stars, why moons orbit planets such as Mars, and the behavior of satellites.
Centripetal Force
Centripetal force is the force required to keep an object moving in a circular path and is directed towards the center around which the object is moving. Its magnitude is given by the mass of the object, the speed of the object, and the radius of the circular path. The equation for centripetal force is \[F = \frac{mv^2}{r}\] where,
  • \(F\) is the centripetal force,
  • \(m\) is the mass of the object,
  • \(v\) is the velocity of the object,
  • \(r\) is the radius of the circular path.
For a satellite in orbit, such as Mars's moon Phoebus, this centripetal force is provided by the gravitational attraction of the planet. Understanding the relationship between centripetal force and the velocity of the orbiting object is vital in orbital mechanics for predicting the behavior of satellites and other celestial bodies.
Orbital Mechanics
Orbital mechanics is a branch of physics that deals with the movement of spacecraft and celestial bodies under the influence of forces such as gravity. It involves the application of Newton's laws of motion and the law of Universal Gravitation to predict the paths of objects as they travel through space.

In our context, we're looking at Phoebus, a satellite of Mars, where we apply principles of orbital mechanics to determine the mass of Mars from characteristics of Phoebus's orbit. Important aspects include the orbital period, \(T\), and the radius, \(r\), of the orbit. The speed of the satellite, \(v\), is given by: \[v = \frac{2\pi r}{T}\] Using orbital mechanics we derive the relationship between the mass of Mars and Phoebus’s orbital radius and period to conclude Mars's mass. Understanding these principles is essential for calculating orbits and maneuvers necessary to propel and sustain satellites in space.
Satellite Physics
Satellite physics encompasses the scientific principles governing the behavior of artificial satellites in orbit, including their trajectory, the forces acting upon them, and their interaction with the celestial body they orbit. A key feature of an artificial satellite, like Phoebus, is that it requires a balance between gravitational pull and the centripetal force to achieve a stable orbit.

This balance is a delicate one: too much speed and the satellite might break free from the orbit; too little and it might fall back to Mars. Understanding how to manipulate satellite velocities, orbits, and positions is essential in satellite communications, navigation, and space exploration. The precision of these calculations allows us to maintain satellites in orbit and use them for various applications, including the monitoring of planetary bodies, like the mass determination of Mars in our exercise.

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Most popular questions from this chapter

A dentist's drill starts from rest. After \(3.20 \mathrm{~s}\) of constant angular acceleration, it turns at a rate of \(2.51 \times\) \(10^{4} \mathrm{rev} / \mathrm{min}\). (a) Find the drill's angular acceleration. (b) Determine the angle (in radians) through which the drill rotates during this period.

In Robert Heinlein's The Moon Is a Harsh Mistress, the colonial inhabitants of the Moon threaten to launch rocks down onto Earth if they are not given independence (or at least representation). Assuming a gun could launch a rock of mass \(m\) at twice the lunar escape speed, calculate the speed of the rock as it enters Earth's atmosphere.

A \(40.0\)-kg child takes a ride on a Ferris wheel that rotates four times each minute and has a diameter of \(18.0 \mathrm{~m}\). (a) What is the centripetal acceleration of the child? (b) What force (magnitude and direction) does the seat exert on the child at the lowest point of the ride? (c) What force does the seat exert on the child at the highest point of the ride? (d) What force does the seat exert on the child when the child is halfway between the top and bottom?

(a) A luggage carousel at an airport has the form of a section of a large cone, steadily rotating about its vertical axis. Its metallic surface slopes downward toward the outside, making an angle of \(20.0^{\circ}\) with the horizontal. A \(30.0-\mathrm{kg}\) piece of luggage is placed on the carousel, \(7.46 \mathrm{~m}\) from the axis of rotation. The travel bag goes around once in \(38.0 \mathrm{~s}\). Calculate the force of static friction between the bag and the carousel. (b) The drive motor is shifted to turn the carousel at a higher constant rate of rotation, and the piece of luggage is bumped to a position \(7.94 \mathrm{~m}\) from the axis of rotation. The bag is on the verge of slipping as it goes around once every \(34.0 \mathrm{~s}\). Calculate the coefficient of static friction between the bag and the carousel.

Q|C A bicycle is turned upside down while its owner repairs a flat tire. A friend spins the other wheel and observes that drops of water fly off tan- gentially. She measures the heights reached by drops moving vertically (Fig. P7.8). A drop that breaks loose from the tire on one turn rises vertically \(54.0\) cm above the tangent point. A drop that breaks loose on the next turn rises \(51.0 \mathrm{~cm}\) above the tan- gent point. The radius of the wheel is \(0.381 \mathrm{~m}\). (a) Why does the first drop rise higher than the second drop? (b) Neglecting air friction and using only the observed heights and the radius of the wheel, find the wheel's angular acceleration (assuming it to be constant).

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