Chapter 13.3, Problem 34PPSE

Solution

i.

To find: The distance traveled by satellite in given hours.

The distance covered in $1,3,2.5$ and $25$ hours are $11048,33144,27620$ and $276198\text{ km}$ .

Given:

The radius of the earth is $6400\text{ km}$ and the position of the satellite is $35,800\text{ km}$ above the earth’s surface.

Calculation:

The radius of the satellite is $35800+6400=42200\text{ km}$ . So,

The circumference is $2\pi \times 42200\approx 265150\text{ km}$

The satellite takes $24\text{ hours}$ to complete one cycle. So, the speed of the satellite is,

  $\begin{array}{c}\text{speed}=\frac{\text{distance}}{\text{time}}\\ =\frac{265150}{24}\\ =11047.9\text{km}/\text{hr}\end{array}$

So, the distance covered by satellite in $1$ hour is,

  $1\times 11047.9\approx 11048\text{ km}$

Similarly, the distance covered by satellite in $3$ hour is,

  $3\times 11047.9\approx 33144\text{ km}$

And the distance covered by satellite in $2.5$ hour is,

  $2.5\times 11047.9\approx 27620\text{ km}$

Finally, the distance covered by satellite in $25$ hour is,

  $25\times 11047.9\approx 276198\text{ km}$

Conclusion:

Hence, the distance covered in $1,3,2.5$ and $25$ hours are $11048,33144,27620$ and $276198\text{ km}$ .

ii.

To find: The number of hours taken by satellite to travel $200,000$ km.

  $18.1\text{ hours}$

Given:

The distance is $200,000$ km.

Concept used:

To compute the time, use the formula $\text{speed}=\frac{\text{distance}}{\text{time}}$ .

Calculation:

It is understood that the speed of satellite is $11047.9\text{km}/\text{hr}$ .

So, the time taken to travel $200,000$ km is,

  $\begin{array}{c}\text{time}=\frac{\text{distance}}{\text{speed}}\\ =\frac{200,000}{11047.9}\\ =18.1\text{ hours}\end{array}$

Conclusion:

The time taken by the satellite is $18.1\text{ hours}$ .