Question

if the diameter of earth is increases 4 times then the critical velocity becomes??​

Answer 1

Answer:

Well, the question is not well-formed; “critical velocity” is not a physical concept. You may mean “orbital velocity” or “escape velocity”. I could answer either of those questions, but you haven’t added one critical piece of information: does the mass of the Earth stay the same, or does it increase with the increase in radius? In other words, does the density stay the same? That said, let’s do the calculations under both assumptions.

Orbital velocity at distance r is given by

v2r=GMr2

or

v=GMr−−−√

Let’s write the current orbital velocity as vo , the new velocity as vn , the current mass as Mo , new mass as Mn , current radius as ro , and the new radius as rn , and you have given us that rn=2ro . So

vn=GMnrn−−−−√=GMn2ro−−−−√

vo=GMoro−−−−√

And you want us to calculate vnvo

Dividing

vnvo=GMnroGMo2ro−−−−−−√=Mn2Mo−−−√

So (1), if the radius doubles but mass stays the same, then

vnvo=12−−√

But if the density stays the same, well, then

M=4πdr33

So

MnMo=12πdr3n12πdr3o=8r3or3o=8

So

vnvo=Mn2Mo−−−√=82−−√=4–√=2

The orbital velocity would double.

Escape velocity is always 2–√ orbital velocity, so we can multiply the answer for orbital velocity.

In the case of the Earth, orbital velocity at the surface is about 7.9 km/s, so if the radius doubled, we have the following:

(1) No change in mass. Orbital velocity is 5.6 km/s, escape velocity is 7.9 km/s

(2) No change in density. Orbital velocity is 15.8 km/s, escape velocity is 22.3 km/s