Question

for an object lying on the surface of the earth the escape speed is 11.2 km per second if an object on the earth is thrown away with speed twice times this value find its speed after it escaped from the gravitational field of the earth​

Answer 1

Answer:

Explanation:

Here escape velocity of the earth is 11.2 km/s

and that object is thrown with 22.4 km/s

so to find final velocity

here gravitational field is conservative force so we can use energy conservation

and gravtitational potetial energy of the projectile far away from the earth is 0

so,  

1/2 mv² (f) -1/2 mv²(i)= 1/2 mv²(final)  

1/2 m will get cancelled

22.4²-11.2²=v²  

so v will be approximately equal to 19.3 m/s