Question

The orbital period of a satellite is 2 × 106 s and its total radius is 2.5 × 1012 m. The tangential speed of the satellite, written in standard notation, is ____ m/s.

Answer 1

Answer:2.5π*10⁶ m/s

Explanation:

frequency=1/time period

time period is given = 2*10⁶ s

so frequency will be = 5*10⁻⁷ Hz

so angular frequency (ω) will be π*10⁻⁶ [ω =2π*frequency]

V(tangential)= ωr

=π*10⁻⁶*2.5*10¹² = 2.5π*10⁶ m/s

Answer 2

The orbital period of the satellite[T] is given as $2*10^{6} S$.

The radius of the satellite is given [R] $2.5*10^{12} m$.

we are asked here to calculate the tangential speed of the satellite.

Before going to get the solution first we have understand the tangential speed.

The tangential speed of a satellite is given as the speed required to keep the satellite along the orbit. If satellite speed is less than tangential speed,there is the chance of it falling down towards earth. If it is more,then it will deviate from it orbit and can't stick to the orbit further.In a simple way  the tangential speed is the linear speed of an object in a circular path.

Now we have to calculate the tangential speed [V].

Mathematically the tangential speed [V]   written as -

                                V$=\frac{2\pi R}{T}$

where T is the time period of the satellite and R is the radius of the satellite.

                        $V=\frac{2*3.14*10^{12} }{2*10^{6} }$

                               $= 7.85*10^{6} m/s$

There is also another way through which we can get  the solution as explained below-

We know that the tangential speed of a satellite V$=\sqrt{\frac{GM}{R^{2} } }$

where G is the gravitational constant and M is the mas of central object.

But we know that $g=\frac{GM}{R^{2} }$

                               ⇒$GM=gR^{2}$  where g is the acceleration due to gravity of that central object.

Hence    $V=\sqrt{\frac{gR^{2} }{R} }$

               ⇒   $V=\sqrt{gR}$

By knowing the value of g due to that central object we can also calculate its tangential speed.