Question

A satellite is revolving around the earth in circular orbit.If the radius of the orbit is increased from R to 2R,What will be its velocity​

Answer 1

Explanation:

Ignoring orbital decay, gravitational aberrations, etc., the satellite at 2R travels at .7 the velocity of the satellite at R, derived using the standard orbital formula:

Vr=(Ge∗Me)/r−−−−−−−−−−√

where:

r= the distance of the satellite to the center of Earth.

Ge = gravitational constant for Earth (6.673e-11 Nm^2/kg^2), and

Me = Mass of Earth (5.98e24kg)

Therefore, the velocity at radius R, call it Vr, would be:

Vr=(6.673e−11∗5.98e24)/R−−−−−−−−−−−−−−−−−−√=3.99e14/R−−−−−−−−√

Similarly, the velocity at 2R (V2r) would be:

V2r=(6.673e−11∗5.98e24)/2R−−−−−−−−−−−−−−−−−−−√=3.99e14/2R−−−−−−−−−√

Intuitively, we can see that V2r=Vr/1.4=.7Vr

I believe the answer you’re looking for is V2r travels at .7 the velocity of the satellite at Vr

Answer 2

Answer:

Explanation:

Explanation:

Ignoring orbital decay, gravitational aberrations, etc., the satellite at 2R travels at .7 the velocity of the satellite at R, derived using the standard orbital formula:

Vr=(Ge∗Me)/r−−−−−−−−−−√

where:

r= the distance of the satellite to the center of Earth.

Ge = gravitational constant for Earth (6.673e-11 Nm^2/kg^2), and

Me = Mass of Earth (5.98e24kg)

Therefore, the velocity at radius R, call it Vr, would be:

Vr=(6.673e−11∗5.98e24)/R−−−−−−−−−−−−−−−−−−√=3.99e14/R−−−−−−−−√

Similarly, the velocity at 2R (V2r) would be:

V2r=(6.673e−11∗5.98e24)/2R−−−−−−−−−−−−−−−−−−−√=3.99e14/2R−−−−−−−−−√

Intuitively, we can see that V2r=Vr/1.4=.7Vr

I believe the answer you’re looking for is V2r travels at . the velocity of the satellite at Vr