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Chapter 13: Problem 90

A spacecraft is describing a circular orbit at an altitude of \(1500 \mathrm{km}\) above the surface of the earth. As it passes through point \(A,\) its speed is reduced by 40 percent and it enters an elliptic crash trajectory with the apogee at point \(A\). Neglecting air resistance, determine the speed of the spacecraft when it reaches the earth's surface at point \(B .\)

Short Answer

Expert verified
The spacecraft's speed at Earth's surface is approximately 10950 m/s.

Step by step solution

01

Understand initial conditions

First, calculate the initial velocity when the spacecraft is in a circular orbit. Since it's 1500 km above Earth's surface, the radius of the orbit is the Earth's radius (6371 km) plus 1500 km, which is 7871 km or 7871000 m. Use the formula for orbital speed: \[ v_i = \sqrt{\frac{GM}{r}} \]where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \text{ m}^3\text{/kg/s}^2 \) and \( M \) is Earth's mass (\( 5.972 \times 10^{24} \text{ kg} \)). Plug in values to find \( v_i \).
02

Calculate speed after reduction

The spacecraft's speed is reduced by 40%, so its new speed \( v_a \) at point A is 60% of the initial speed. Therefore, \[ v_a = 0.6 \times v_i \]. Calculate \( v_a \) using \( v_i \) from Step 1.
03

Determine semi-major axis of the ellipse

Since point A is the apogee of the elliptical trajectory, use the fact: \[ v_a^2 = GM \left(\frac{2}{r_a} - \frac{1}{a}\right) \]where \( r_a \) is the apogee distance (7871000 m) and \( a \) is the semi-major axis. Rearrange the formula to solve for \( a \).
04

Find speed at Earth's surface

The perigee (point B) is at Earth's surface (r = 6371000 m). Use the equation for velocity at any point in an elliptical orbit:\[ v_b = \sqrt{GM\left(\frac{2}{r_b} - \frac{1}{a}\right)} \]where \( r_b \) is the Earth's radius. Calculate \( v_b \) using \( a \) from Step 3.

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Key Concepts

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

Elliptical Trajectory
When a spacecraft orbits around a celestial body like Earth, its trajectory can form different shapes, such as circular or elliptical paths. An elliptical trajectory is one where the path resembles a flattened circle, or an oval. This is the common shape for most planetary orbits and is defined by two foci.
In the context of the exercise, after the spacecraft's speed was reduced at point A, it entered an elliptical trajectory with the apogee at point A itself, which is the furthest point in this new orbit from Earth's center. The closest point to Earth during this trajectory is called the perigee.
Key points about elliptical orbits:
  • Elliptical orbits are defined by their major and minor axes, with the semi-major axis defining half of the longest diameter of the ellipse.
  • The shape and size of the ellipse depend on the energy and angular momentum of the spacecraft.
  • Elliptical orbits have varying speeds, with the spacecraft moving fastest at the perigee and slowest at the apogee.
Orbital Speed
Understanding orbital speed is crucial in orbital mechanics, particularly when dealing with spacecraft orbiting Earth. Orbital speed is the velocity needed for an object to orbits a celestial body.
In a circular orbit, such as the initial path of the spacecraft, orbital speed ensures that the centripetal force needed for circular motion is provided by gravity.
For a circular orbit, the formula is: \[ v = \sqrt{\frac{GM}{r}} \]where:
  • \( v \) is the orbital speed,
  • \( G \) is the gravitational constant,
  • \( M \) is the mass of the Earth, and
  • \( r \) is the radius of the orbit.
In the example provided, we used this formula to calculate the initial speed of the spacecraft before it was reduced by 40%. This change set it on a different trajectory, changing its distance speed across the orbit.
Gravitational Constant
The gravitational constant, denoted by \( G \), is a key component in the equations of orbital mechanics. It quantifies the intensity of gravity between any two masses.
With a value of \( 6.674 \times 10^{-11} \ \text{m}^3/\text{kg/s}^2 \), \( G \) allows us to calculate forces between celestial bodies.
In the context of the problem, the gravitational constant plays a crucial role in determining the various speeds and distances involved in the trajectories, as it is a part of the orbital speed and energy equations.
\( G \) is important to:
  • Calculate the gravitational force between Earth and the spacecraft.
  • Determine the spacecraft's speed at any point on its orbital trajectory.
  • Understand the energy dynamics throughout the spacecraft's flight path.
Semi-Major Axis
A fundamental aspect of an elliptical orbit is the semi-major axis, which defines half of the ellipse's longest diameter. It is an essential parameter in Kepler's laws of planetary motion, and greatly influences the dynamics of orbital motion.
In calculations, the semi-major axis \( a \) helps determine properties such as orbital period and energy.
In our exercise, after the spacecraft shifted to an elliptical orbit, the semi-major axis was crucial in finding the speed of the spacecraft at different points in its trajectory:
  • We need the semi-major axis (\( a \)) to calculate the speed at the perigee using the formula: \[ v = \sqrt{GM\left(\frac{2}{r} - \frac{1}{a}\right)} \]
  • The semi-major axis helps express energies at different points during the ellipse and design trajectories for orbit insertion and transfer.

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Most popular questions from this chapter

Determine the maximum theoretical speed that may be achieved over a distance of 110 m by a car starting from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75, and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) front-wheel drive, (b) rear-wheel drive.

At an amusement park there are 200 -kg bumper cars \(A, B,\) and \(C\) that have riders with masses of \(40 \mathrm{kg}, 60 \mathrm{kg},\) and \(35 \mathrm{kg}\), respectively. Car \(A\) is moving to the right with a velocity \(\mathrm{v}_{A}=2 \mathrm{m} / \mathrm{s}\) when it hits stationary car \(B\). The coefficient of restitution between each car is \(0.8 .\) Determine the velocity of car \(C\) so that after car \(B\) collides with car \(C\) the velocity of \(\operatorname{car} B\) is zero.

A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 27 m and the radius of CD is 72 m. The car and its occupants, of total mass 250 kg, reach point A with practically no velocity and then drop freely along the track. Determine the maximum and minimum values of the normal force exerted by the track on the car as the car travels from A to D. Ignore air resistance and rolling resistance.

A package is projected up a \(15^{\circ}\) incline at \(A\) with an initial velocity of \(8 \mathrm{m} / \mathrm{s}\). Knowing that the coefficient of kinetic friction between the package and the incline is \(0.12,\) determine \((a)\) the maximum distance \(d\) that the package will move up the incline, \((b)\) the velocity of the package as it returns to its original position.

Ball \(B\) is hanging from an inextensible cord. An identical ball \(A\) is released from rest when it is just touching the cord and drops through the vertical distance \(h_{A}=8\) in. before striking ball \(B .\) Assuming \(e=0.9\) and no friction, determine the resulting maximum vertical displacement \(h_{B}\) of the ball \(B .\)
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