/*! 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 104 As a first approximation to the ... [FREE SOLUTION] | 91Ó°ÊÓ

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As a first approximation to the analysis of a space flight from the earth to Mars, it is assumed that the orbits of the earth and Mars are circular and coplanar. The mean distances from the sun to the earth and to Mars are \(149.6 \times 10^{6} \mathrm{km}\) and \(227.8 \times 10^{6} \mathrm{km}\), respectively. To place the spacecraft into an elliptical transfer orbit at point \(A,\) its speed is increased over a short interval of time to \(v_{A},\) which is faster than the earth's orbital speed. When the spaceraft reaches point \(B\) on the elliptical transfer orbit, its speed \(v_{B}\) increased to the orbital speed of Mars. Knowing that the mass of the sun is \(332.8 \times 10^{3}\) times the mass of the earth, determine the increase in velocity required ( \(a\) ) at \(A,(b)\) at \(B\).

Short Answer

Expert verified
Increase in velocity: \(2.94 \text{ km/s}\) at A and \(2.65 \text{ km/s}\) at B.

Step by step solution

01

Find Earth's Orbital Speed

The orbital speed of a planet in its circular path is given by the formula,\[v = \sqrt{\frac{G M_s}{r}}.\]Here, the gravitational constant \(G\) and solar mass \(M_s\) can be merged given \(\frac{G M_s}{r} = \frac{4\pi^2 r}{T^2}\). For Earth's orbital speed where \(r = 149.6 \times 10^6 \). Substituting known values into\[v_e = \sqrt{\frac{G 332.8 \times 10^3 M_e}{149.6 \times 10^6}}.\] where \(G\) is the universal gravitational constant, \(M_e\) is Earth's mass. Simplifying, the Earth's orbital speed \(v_e\) is approximately \(29.78 \text{ km/s}\).
02

Calculate Speed at Perihelion (Point A)

To determine the speed increase at point A, apply the vis-viva equation for point A of the elliptical transfer orbit: \[v_A = \sqrt{G M_s \left(\frac{2}{r_1} - \frac{1}{a}\right)}\]where \(r_1 = 149.6 \times 10^6 km\), and semi-major axis, \(a = \frac{(r_1 + r_2)}{2} = \frac{(149.6 + 227.8) \times 10^6}{2}\). Calculating,\[a =189.7 \times 10^6 \text{ km}\]. Substituting back, we find \(v_A\approx 32.72 \text{ km/s}\).
03

Find Velocity Change at A

Determine the increase in velocity at point A. The change \(\Delta v_A\) is given by:\[\Delta v_A = v_A - v_e\].Substituting the values calculated for \(v_A = 32.72 \text{ km/s}\) and \(v_e = 29.78 \text{ km/s}\), we get,\[\Delta v_A \approx 2.94 \text{ km/s}.\]
04

Calculate Speed at Aphelion (Point B)

Use the vis-viva equation for point B of the elliptical transfer orbit: \[v_B = \sqrt{G M_s \left(\frac{2}{r_2} - \frac{1}{a}\right)}\]where \(r_2 = 227.8 \times 10^6 km\).Substituting the known values, we find \(v_B = 21.48 \text{ km/s}\).
05

Calculate Mars' Orbital Speed

Using the orbital speed formula at point B (Mars' orbit): \[v_m = \sqrt{\frac{G M_s}{r_2}}\].Substituting \(r_2 = 227.8 \times 10^6\) gives \(v_m \approx 24.13 \text{ km/s}\).
06

Find Velocity Change at B

The increase in velocity at point B is\(\Delta v_B = v_m - v_B\):\[\Delta v_B = 24.13 - 21.48 = 2.65 \text{ km/s}.\]

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Key Concepts

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

Earth's Orbital Speed
Understanding Earth's orbital speed is crucial in predicting the motion of any spacecraft departing from our planet. The Earth's orbital speed is determined by a combination of factors, most notably the gravitational force exerted by the sun and the radius of Earth's orbit. In this context, the formula relevant is:
  • \[ v = \sqrt{\frac{GM_s}{r}} \]
Here, \(G\) is the gravitational constant, \(M_s\) is the mass of the sun, and \(r\) is the average distance from the Earth to the sun. This formula highlights how the speed at which Earth orbits doesn’t change randomly but is a direct result of these defined parameters.
For Earth, substituting known values like the solar mass (considerably larger than Earth's mass) simplifies to an orbital speed approximately around 29.78 km/s. This speed ensures Earth's stable orbit around the sun.
Vis-Viva Equation
The vis-viva equation is an essential tool in orbital mechanics. It gives us the speed of an object in an elliptical orbit at any given point. Specifically, it is:
  • \[ v = \sqrt{G M_s \left(\frac{2}{r} - \frac{1}{a}\right)} \]
This calculation involves:
  • \(r\): the distance of the object from the center of the mass it orbits (in our case, Earth or Mars from the sun).
  • \(a\): the semi-major axis of the orbit.

To put this into practical terms, when a spacecraft departs Earth towards Mars, it initially transitions into an elliptical transfer orbit. Its speed can be predicted at any point using this equation. At point A, for instance, adjustments to the spacecraft's speed are needed to enter this transfer orbit, calculated through vis-viva, showing an increase in speed over the Earth's orbital speed. Similarly, at point B, further adjustments ensure alignment with Mars’ orbit.
Velocity Change in Space Flight
In space flight, understanding velocity change or \(\Delta v\) is fundamental for mission planning. Velocity change (\(\Delta v\)) describes how much the spacecraft needs to speed up or slow down to transition between different orbits.
When initiating a transfer from Earth to Mars:
  • At point A, the spacecraft speeds up to enter the transfer orbit, requiring a calculated speed increase over Earth's orbital speed.
  • At point B, upon reaching the vicinity of Mars, another speed adjustment aligns it with Mars’ orbit.

These speed changes are critical for efficient propulsion systems. They determine the amount of fuel needed and are crucial for the mission's economic viability. For our example journey to Mars, the spacecraft requires a \(\Delta v\) of approximately 2.94 km/s at point A and about 2.65 km/s at point B. By accurately calculating these, the transfer can be effectively executed with minimal energy expense.

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Most popular questions from this chapter

At an amusement park there 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 \(\mathrm{v}_{\mathrm{A}}=2 \mathrm{m} / \mathrm{s}\) and \(\mathrm{car} C\) has a velocity \(\mathrm{v}_{B}=1.5 \mathrm{m} / \mathrm{s}\) to the left, but car \(B\) is initially at rest. The coefficient of restitution between each car is \(0.8 .\) Determine the final velocity of each car, after all impacts, assuming (a) cars \(A\) and \(C\) hit car \(B\) at the same time, \((b)\) car \(A\) hits car \(B\) before car \(C\) does.

A satellite is projected into space with a velocity \(\mathbf{v}_{0}\) at a distance \(r_{0}\) from the center of the earth by the last stage of its launching rocket. The velocity \(\mathbf{v}_{0}\) was designed to send the satellite into a circular orbit of radius \(r_{0}\). However, owing to a malfunction of control, the satellite is not projected horizontally but at an angle \(\alpha\) with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite.

Observations show that a celestial body traveling at \(1.2 \times 10^{6} \mathrm{mi} / \mathrm{h}\) appears to be describing about point \(B\) circle of radius equal to 60 light years. Point \(B\) is suspected of being a very dense concentration of mass called a black hole. Determine the ratio \(M_{B} / M_{S}\) of the mass at \(B\) to the mass of the sun. (The mass of the sun is \(330,000\) times the mass of the earth, and a light year is the distance traveled by light in 1 year at \(186,300 \mathrm{mi} / \mathrm{s}\).)

A truck is traveling down a road with a 4-percent grade at a speed of 60 mi/h when its brakes are applied to slow it down to 20 mi/h. An antiskid braking system limits the braking force to a value at which the wheels of the truck are just about to slide. Knowing that the coefficient of static friction between the road and the wheels is 0.60, determine the shortest time needed for the truck to slow down.

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.

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