Question

a LA to B
use
A artificial sateliite, is moving in circular
Orbit of radius 4 2250km
calculate its
speed it it itakes 24 hrs to mevolue auound
the earth.

Answer 1

Proper question: A artificial satellite is moving in circular orbit of radius 42250 km, calculate its speed if it takes 24 hrs to revolve around the earth.

Provided that:

• Radius of circular orbit = 42250 kilometres.
• Time taken = 24 hours

To calculate:

• The speed

Solution:

• The speed = 11055.41 km/h

Using concepts:

• Circumference of circle formula
• Speed formula

Using formulas:

• ${\small{\underline{\boxed{\sf{Circumference \: = 2 \pi r}}}}}$

• ${\small{\underline{\boxed{\sf{Speed \: = \dfrac{Distance}{Time}}}}}}$

Explanation: As it's given that the satellite is moving in a circular orbit, therefore, the distance became the circumference of the given circular path. So at the place of distance we have to use circumference formula. Let's see how to solve!

Required solution:

$:\implies \sf Speed \: = \dfrac{Distance}{Time} \\ \\ :\implies \sf Speed \: = \dfrac{Circumference \: of \: circle}{Time} \\ \\ :\implies \sf Speed \: = \dfrac{2 \pi r}{Time} \\ \\ :\implies \sf Speed \: = \dfrac{2 \times 3.14 \times 42250}{24} \\ \\ :\implies \sf Speed \: = \dfrac{6.28 \times 42250}{24} \\ \\ :\implies \sf Speed \: = \dfrac{265330}{24} \\ \\ :\implies \sf Speed \: = 11055.41 \: kmh^{-1}$

Answer 2

Answer:-

• Radius of the circular orbit=r=42250km
• Time=24hr

Finding circumference of orbit:-

$\boxed{\sf Circumference=2\pi r}$

$\\ \rm\longmapsto Circumference=2\times \dfrac{22}{7}\times 42250$

$\\ \rm\longmapsto Circumference =\dfrac{1859000}{7}$

$\\ \rm\longmapsto Circumference=265571.4km$

Now

$\boxed{\sf Speed=\dfrac{Distance}{Time}}$

$\\ \rm\longmapsto Speed=\dfrac{265571.4}{24}$

$\\ \rm\longmapsto Speed=11065km/h(Approx)$

Note:-

Don't be confused by seeing different answers.

• There are Different values of $\Pi$are used .
• In my calculations I used it as 22/7 .

Some other values of pi are:-

• 3.14
• 3.141
• $\rm \sqrt{10}$