Fundamentals of Applied Electromagnetics (7th Edition)
Fundamentals of Applied Electromagnetics (7th Edition)
7th Edition
ISBN: 9780133356816
Author: Fawwaz T. Ulaby, Umberto Ravaioli
Publisher: PEARSON
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Chapter 10, Problem 1P

A remote sensing satellite is in circular orbit around Earth at an altitude of 1,100 km above Earth’s surface. What is its orbital period?

Expert Solution & Answer
Check Mark
To determine

The orbital period.

Answer to Problem 1P

The orbital period is 107.22minutes.

Explanation of Solution

Given data:

The altitude of satellite from Earth’s surface is 1,100km.

Calculation:

The orbit radius R0 is written as,

R0=Re+h (1)

Here,

Re is Earth’s radius, which has standard value as (6,378km).

h is the height of satellite in orbit from the Earth’s surface.

Substitute 6,378km for Re, and 1,100km for h in equation (1).

R0=6,378km+1,100km=7,478km

The conversion of km into m is done as,

1km=1×103m

Hence, the conversion of 7,478km into km is,

7,478km=7,478×103m

The orbital period T is written as,

T=(4π2R03GMe)12 (2)

Here,

T is the orbital period.

G is the gravitational constant.

Me is the Earth mass.

Substitute 7478×103m for R0, 6.67×1011N-m2/kg2 for G, and 5.98×1024kg for Me in equation (2).

T=(4×π2×(7478×103m)3(6.67×1011N-m2/kg2)×5.98×1024kg)12=6433.45seconds

The conversion of seconds into minutes is done as,

1second=160minutes

Hence, the conversion of seconds into minutes is done as,

6433.45seconds=6433.4560minutes=107.22minutes

Conclusion:

Therefore, the orbital period is 107.22minutes.

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