wiz-icon
MyQuestionIcon
MyQuestionIcon
2
You visited us 2 times! Enjoying our articles? Unlock Full Access!
Question

A planet has radius equal to 2RE and density is equal to 4ρE. If the value of escape speed at surface of planet is α times vE (vE is escape speed on earth, RE is radius of earth and ρE is the density of earth) then -

A
α=2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
α=9
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
α=6
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
α=4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D α=4
vE=2GMERE=2G×ρE×4πR3E3RE
vE=8πGρER2E3

If radius is 2RE and density is 4ρE then,
vE=8πG(4ρE)(2RE)23
vE=16×8πGρER2E3
vE=4vE

Therefore, α=4.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Escape Velocity
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon