AP EAPCET 08-Jul-2022 Shift 1 Paper

Given, v= escape speed
So, vescape =v=√

GM

R

×2 ...(i)
M= mass of earth,
R= Radius of earth
Now, when an object is projected with speed 4v from surface of earth, let its speed at infinity is v′

Then by energy conservation,
Initial energy = final energy
⇒−

GMm

R

+

1

2

m(4v)2=0+

1

2

m(v′)2
⇒

−GMm

R

+16(

1

2

mv2)=

1

2

m(v)2 ...(ii)
Substituting value of v from Eq. (i) in Eq. (ii), we get

−GMm

R

+16(

1

2

m

GM×2

R

)=

1

2

m(v′)2
⇒

15GMm

R

=

1

2

m(v′)2
⇒v′2=15×

2GM

R

⇒v′2=15×v2 ...[∴v2=‌

2GM

R

]
⇒v′=√15⋅v