TS EAMCET 04 May 2019 Shift 2 Solved Paper

The escape velocity of the surface of earth is given by,
ve=√2gRe
Substitute 6400km for Re and 10m∕s2 for g in equation (I).
ve‌‌=√2(10m∕s2)(6400km×‌

103m

1km

)
‌‌=11.3×103m∕s
‌‌=11.3km∕s
‌‌≈12km∕s
When
g′=2g
And
R′=‌

Re

2

The escape velocity is calculated as,

ve′=√2((2)10m∕s2)(‌

6400km× 103m 1km

2

)

=11.3×103m∕s
=11.3km∕s
≈12km∕s