CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A satellite is to be placed in equatorial geostationary orbit around earth for communication. The height of such a satellite is [ME=6×1024kg,RE=6400km,T=24h,G=6.67×1011Nm2kg2]

A
3.57×105m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3.57×106m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
3.57×107m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
3.57×108m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 3.57×107m
Time period of satellite, T=2π(RE+h)GME(RE+h)=2π(RE+h)3/2GME
Squaring both sides, we get T2=4π(RE+h)3GME
(RE+h)3=GMET24π2
(RE+h)=(GMET24π2)1/3
or h=(GMET24π2)1/3RE
Here, ME=6×1024kg
RE=6400km=6400×103m=6.4×106m
T=24h=24×60×60s=86400s
G=6.67×1011Nm2kg2
On substituting the given values, we get
h=(6.67×1011×6×1024×(86400)24×(3.14)2)6.4×106
=4.21×1076.4×106=3.57×107m

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