Question

Consider a satellite in a circular orbit around the Earth. If it were at an altitude equal to twice the radius of the Earth, 2RE, how would its speed v relate to the Earth's radius RE, and the magnitude g of the acceleration due to gravity on the Earth's surface?

Answer 1

Answer:

$v=\sqrt{\frac{gR_E}{2}}$

Explanation:

Satellites experiment a force given by Newton's Gravitation Law:

$F=\frac{GMm}{r^2}$

where M is Earth's mass, m the satellite's mass, r the distance between their gravitational centers and G the gravitational constant.

We also know from Newton's 2nd Law that F=ma, so putting both together we will have:

$ma=\frac{GMm}{r^2}$

$a=\frac{GM}{r^2}$

If we are on the surface of the Earth, the acceleration would be g and $r=R_E$ (Earth's radius):

$g=\frac{GM}{R_E^2}$

Which we will write as:

$gR_E^2=GM$

If we are on orbit the acceleration is centripetal ($a=\frac{v^2}{r}$), so we have:

$\frac{v^2}{r}=a=\frac{GM}{r^2}=\frac{gR_E^2}{r^2}$

$v^2=\frac{gR_E^2}{r}$

$v=\sqrt{\frac{gR_E^2}{r}}$

And if this orbit has a radius $r=2R_E$ we have:

$v=\sqrt{\frac{gR_E^2}{2R_E}}=\sqrt{\frac{gR_E}{2}}$

Answer 2

Answer:

The relation of speed v with the “RE” and “g” will be

                    v= √((gRE)/2)  

Explanation:

When a satellite is orbiting around the earth, the mathematical relation of its speed v is:

                        v= √((gRE^2)/r)    …….. (i)

Where,

g = gravitational acceleration  

RE = radius of earth  

r = hight from the surface of earth

In the question it is given that,  

                          r = 2RE              ……… (ii)

Putting equation (ii) in (i), we get  

                         v= √((gRE)/2)