/*! 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 119 The satellite \(S\) travels arou... [FREE SOLUTION] | 91Ó°ÊÓ

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The satellite \(S\) travels around the earth in a circular path with a constant speed of \(20 \mathrm{Mm} / \mathrm{h}\). If the acceleration is \(2.5 \mathrm{~m} / \mathrm{s}^{2}\), determine the altitude \(h\). Assume the earth's diameter to be \(12713 \mathrm{~km}\).

Short Answer

Expert verified
The altitude of the satellite from the surface of the Earth is approximately \(1228774.736 \mathrm{km}\).

Step by step solution

01

Convert values to common units

To solve this problem, first convert the given values to common units. Speed of \(20 \mathrm{Mm} / \mathrm{h}\) is converted into meters per second (m/s) and acceleration remains in meter per seconds squared (m/s\(^2\)). \(20 \mathrm{Mm} / \mathrm{h} = 5555.56 \mathrm{m} / \mathrm{s}\)
02

Apply formula for acceleration in circular motion

Next, apply the formula for acceleration in a circular motion: \(a = v^2 / r\). Here, \(v\) is the velocity (or speed) of the satellite and \(r\) is the radius of the satellite's circular path. From this equation we will solve for the radius \(r\).
03

Solve for radius

Rearranging the formula from step 2, we get \(\ r = v^2 / a\). Substitute the given values \(v = 5555.56 \mathrm{m} / \mathrm{s}\) and \(a = 2.5 \mathrm{m} / \mathrm{s}^{2}\) into the formula to calculate \(r\). Therefore, \(r = (5555.56)^2 / 2.5 = 1235131236 \mathrm{m}\).
04

Calculate altitude

We calculated the radius of the satellite's orbit around the earth. However, we need to find the altitude. The altitude is the distance from the earth's surface to the satellite. As we've calculated the satellite's distance from the center of the earth, we can subtract the earth's radius to find the altitude. We're given that the diameter of the earth is \(12713 \mathrm{km}\), so the radius is \(\frac{12713}{2} = 6356.5 \mathrm{km}\) or \(6356.5 * 1000 = 6356500 \mathrm{m}\). Subtract this from the orbital radius to find the altitude. Hence, \(h = r - \mathrm{earth\_radius} = 1235131236 \mathrm{m} - 6356500 \mathrm{m} = 1235131236 \mathrm{m} - 6356500 \mathrm{m} = 1228774736 \mathrm{m}\).
05

Convert meters to kilometers

Convert the altitude from meters to kilometers to make it easier to understand. \(h = 1228774736 / 1000 = 1228774.736 \mathrm{km}\).

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Key Concepts

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

Uniform Circular Motion
Understanding uniform circular motion is critical for analyzing the movement of satellites around Earth. This type of motion describes an object traveling along a circular path with a constant speed. It's important to note that while the speed remains the same, the direction of the object's velocity is continuously changing. This is due to the object being in a constant state of acceleration towards the center of the circular path, even if it maintains a constant speed in terms of magnitude.

For a satellite like the one in the exercise, which moves at a speed of 20 Mm/h, its motion can be described as uniform circular motion because its speed is constant. Yet, despite this consistency in speed, the satellite is accelerating towards the Earth's center, allowing it to maintain its circular orbit.
Centripetal Acceleration
Centripetal acceleration is at the heart of making circular motion possible. It is the acceleration that keeps an object moving in a circular path, acting inward towards the center of the rotation. It's crucial to understand that this acceleration doesn't change the object's speed but its direction. The formula to calculate centripetal acceleration is given by a = v2 / r, where a is the centripetal acceleration, v is the constant speed of the object, and r is the radius of the circular path.

In the scenario with our satellite, we know the centripetal acceleration is 2.5 m/s2. By applying this concept with the formula, we can determine necessary properties of the satellite's orbit, such as the radius, which is essential to finding the altitude.
Conversion of Units
Being adept at converting units is indispensable in physics and engineering problems, especially when dealing with satellite motion calculations. In this case, we converted the satellite’s speed from megameters per hour (Mm/h) to meters per second (m/s) to align with the SI unit for acceleration (meters per second squared, m/s2).

Converting from Mm/h to m/s requires understanding the relationship between these units: 1 Mm equals 1,000,000 meters and there are 3,600 seconds in an hour. With this knowledge, the conversion becomes a straightforward calculation that is essential for using the formulas correctly and for ensuring consistency in the units when applying physical laws.
Circular Orbit Altitude
The altitude of a satellite in a circular orbit must be calculated from its center-to-center distance from Earth, which we get from the centripetal acceleration formula, minus the Earth's radius. This calculation is vital because it tells us how far above the Earth's surface the satellite is operationally situated.

After calculating the radius of the satellite’s circular orbit, we subtract Earth’s radius from it to find the altitude. As distances in space are astronomically large, they're often presented in kilometers rather than meters for simplicity and ease of comprehension. This altitude is important for various applications, such as determining the satellite's potential orbital period, its visibility from Earth, and its communication range with ground stations.

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Most popular questions from this chapter

Two planes, \(A\) and \(B\), are flying at the same altitude. If their velocities are \(v_{A}=500 \mathrm{~km} / \mathrm{h}\) and \(v_{B}=700 \mathrm{~km} / \mathrm{h}\) such that the angle between their straightline courses is \(\theta=60^{\circ}\), determine the velocity of plane \(B\) with respect to plane \(A\).

For a short time the arm of the robot is extending such that \(\dot{r}=1.5 \mathrm{ft} / \mathrm{s}\) when \(r=3 \mathrm{ft}, z=\left(4 t^{2}\right) \mathrm{ft}\), and \(\theta=0.5 t \mathrm{rad}\), where \(t\) is in seconds. Determine the magnitudes of the velocity and acceleration of the grip \(A\) when \(t=3 \mathrm{~s}\).

The velocity of a particle traveling along a straight line is \(v=v_{0}-k s\), where \(k\) is constant. If \(s=0\) when \(t=0\), determine the position and acceleration of the particle as a function of time.

. A rocket is fired from rest at \(x=0\) and travels along a parabolic trajectory described by \(y^{2}=\left[120\left(10^{3}\right) x\right] \mathrm{m} .\) If the \(x\) component of acceleration is \(a_{x}=\left(\frac{1}{4} t^{2}\right) \mathrm{m} / \mathrm{s}^{2}\) where \(t\) is in seconds, determine the magnitude of the rocket's velocity and acceleration when \(t=10 \mathrm{~s}\).

A block moves outward along the slot in the platform with a speed of \(\dot{r}=(4 t) \mathrm{m} / \mathrm{s}\), where \(t\) is in seconds. The platform rotates at a constant rate of \(6 \mathrm{rad} / \mathrm{s}\). If the block starts from rest at the center, determine the magnitudes of its velocity and acceleration when \(t=1 \mathrm{~s}\).

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