978-0073380308 Chapter 5 Solution Manual Part 23

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Problem 5.142

Explorer 7 was launched on October 13, 1959, with an apogee altitude above the Earth’s surface of

1073 km and a perigee altitude of 573 km above the Earth’s surface. Its orbital period was 101:4 min.

Using this information, calculate Gmefor the Earth and compare it with gr2

Photo credit: NASA

Solution

Recalling that the radius of the Earth

reD6371 km

, we begin with computing the orbit’s radii at perigee and

apogee. Specifically, we have

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Dynamics 2e 1071

Problem 5.143

Explorer 7 was launched on October 13, 1959, with an apogee altitude above the Earth’s surface of

1073 km and a perigee altitude of 573 km above the Earth’s surface. Its orbital period was 101:4 min.

Determine the eccentricity of the Explorer 7’s orbit, as well as its speeds at perigee and apogee.

Photo credit: NASA

Solution

Recalling that the radius of the Earth

reD6371 km

, we begin with computing the orbit’s radii at perigee and

apogee. Specifically, we have

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permission of McGraw-Hill, is prohibited.

1072 Solutions Manual

gD9:81 m=s2

is the acceleration due to gravity on the surface of the Earth, for the speeds at perigee and

apogee we have

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Dynamics 2e 1073

Problem 5.144

Ageosynchronous equatorial orbit is a circular orbit above the Earth’s

equator that has a period of

day (these are sometimes called geostationary

orbits). These geostationary orbits are of great importance for telecommu-

nications satellites because a satellite orbiting with the same angular rate

as the rotation rate of the Earth will appear to hover in the same point in

the sky as seen by a person standing on the surface of the Earth. Using

this information, determine the altitude

and radius

of a geostationary

orbit (in miles). In addition, determine the speed

of a satellite in such an

orbit (in miles per hour).

gr2

e)rgD ⌧2

4⇡2!1=3

To find hgobserve that

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Problem 5.145

An artificial satellite is launched from an altitude of

500 km

with

a velocity

that is parallel to the surface of the Earth. Requiring

that the altitude at apogee be

20;000 km

, determine the velocity

, that is, the position in the orbit when the velocity is first

orthogonal to the launch velocity.

Solution

The radius of Earth is reD6:371⇥106m so

rPDreC500 km )rPD6:871⇥106m;

rADreC20;000 km )rAD26:37⇥106m;

aD1

2.rACrP/)aD16:62⇥106m;

bDprArP)bD13:46⇥106m:

The distance from the center of Earth to Bis

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Dynamics 2e 1075

Problem 5.146

The mass of the planet Jupiter is 318 times that of Earth, and its equatorial radius is

71;500 km

. If a space

probe is in a circular orbit about Jupiter at the altitude of the Galilean moon Callisto (orbital altitude

1:812⇥106km

), determine the change in speed

v

needed in the outer orbit so that the probe reaches a

minimum altitude at the orbital radius of the Galilean moon Io (orbital altitude

3:502⇥105km

). Assume

that the probe is at the maximum altitude in the transfer orbit when the change in speed occurs and that

change in speed is impulsive, that is, it occurs instantaneously.

Solution

Let

and

denote the radii of Callisto’s and Io’s orbits, respectively. The radius

is also the radius of the

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where

is the mass of the Earth. Hence, applying Eq. (4), we can determine the speeds

and

follows:

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Dynamics 2e 1077

Problem 5.147

The on-orbit assembly of the International Space Station (ISS) began

in 1998 and continues today. The ISS has an apogee altitude above

the Earth’s surface of

341:9 km

and a perigee altitude of

331:0 km

above the Earth’s surface. Determine its maximum and minimum

speeds in orbit, its orbital eccentricity, and its orbital period. Re-

search its actual orbital period and compare it with your calculated

value.

Photo credit: NASA

aD1

2.rACrP/D6:707⇥106m:

Hence, using the (full precision value of the) above results, the maximum orbital speed is at perigee and is

given by

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Problem 5.148

The optimal way (from an energy standpoint) to transfer from one

circular orbit about a primary body

to another circular orbit is via

the so-called Hohmann transfer, which involves transferring from one

circular orbit to another using an elliptical orbit that is tangent to both

at the periapsis and apoapsis of the ellipse. The ellipse is uniquely

defined because we know

(the radius of the inner circular orbit) and

(the radius of the outer circular orbit), and therefore we know the

semimajor axis

by Eq. (5.88) and the eccentricity

by Eq. (5.87) or

Eqs. (5.90). Performing a Hohmann transfer requires two maneuvers,

the first to leave the inner (outer) circular orbit and enter the transfer

ellipse and the second to leave the transfer ellipse and enter the outer

(inner) circular orbit.

A spacecraft S1needs to transfer from circular low Earth parking

orbit with altitude

120 mi

above the surface of the Earth to a circular

geosynchronous orbit with altitude

22;240 mi

. Determine the change

in speed

vP

required at perigee

of the elliptical transfer orbit and

the change in speed

vA

required at apogee

. In addition, compute

the time required for the orbital transfer. Assume that the changes in

speed are impulsive, that is, they occur instantaneously.

Solution

Recalling that the radius of the Earth is reD3959 mi D3959.5280/ ft, the radii at perigee and apogee are

Let

be the speed when the spacecraft is in a circular orbit with radius

be the speed at perigee for the

transfer orbit,

be the speed at apogee for the transfer orbit, and

be the speed at the circular orbit with

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where, in addition to using the (full precision) values of

, and

indicated in Eqs. (1)–(3), we have

used

gD32:2 ft=s2

and

reD3959.5280/ ft

. Thus, using the full precision values of

, and

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