two satellites are launched in to cicular orbit of , with radius R and 1.02 around earth . Then percentage difference in time period of satellite is
(1)1%
(2)2%
(3)3/2%
(4)3%
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Answer:
According to Kepler's law,
⟹T
2
∝R
3
⟹T∝R
(3/2)
......................equation (1)
where,
T= time period
R= radius
Now, differentiating both sides we get
dT=(
2
3
)R
(
2
1
)
dR
Now, Dividing both sides by equation (1), we get
⟹
T
dT
=
R
(
2
3
)
(
2
3
)R
(
2
1
)
dR
⟹
T
dT
=
2
3
R
dR
.............................equation (2)
Now,
Percentage difference in radius, =
R
dR
×100%=
R
1.02R−R
×100%=2%
So, Percentage difference in time period , =
T
dT
×100=
2
3
×2=3%
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