A satellite of mass mS orbits the Earth (with mass mE and radius RE ) with a velocity v and an altitude h. The gravitational force FG and the centripetal force FC are given by:
mS ·mE mS ·v2
FG =G(RE +h)2, FC = RE +h, G=const.
(a) Find an equation for the kinetic energy Ekin(h) of a satellite with an altitude h.
(b) Based on the kinetic energy, how much liquid hydrogen (energy density: 106 J/Litre) is at least needed to bring a small 1 kg satellite in an orbit of 400 km. (Use literature to find mE , RE , G.)
Answers
Answer:
Answer:
Given:
Radius of the circular orbits are:
9000 km
9010 km
Period of revolution of faster satellite is 90 minutes.
To find:
Difference in time period of revolution.
Calculation:
According to Kepler's Law of Planetary Motion , we can say that :
{T}^{2} \propto {R}^{3}T
2
∝R
3
For the 1st satellite :
{T_{1}}^{2} \propto {R_{1}}^{3}T
1
2
∝R
1
3
=>{90}^{2} \propto {9000}^{3}=>90
2
∝9000
3
For 2nd satellite :
{T_{2}}^{2} \propto {R_{2}}^{3}T
2
2
∝R
2
3
=>{T_{2}}^{2} \propto {9010}^{3}=>T
2
2
∝9010
3
Dividing the proportionality :
{\dfrac{T_{2}}{T_{1}}}^{2}={\dfrac{9010}{9000}^{3}
=>{\dfrac{T_{2}}{90}}^{2}={\dfrac{9010}{9000}^{3}
=>{(\dfrac{T_{2}}{90})}^{2}=1.0033=>(
90
T
2
)
2
=1.0033
=>\dfrac{T_{2}}{90}= \sqrt{1.0033}=>
90
T
2
=
1.0033
=>\dfrac{T_{2}}{90}= 1.0016=>
90
T
2
=1.0016
=> T_{2} = 90.15 \:min=>T
2
=90.15min
So Difference in time period :
\Delta T = 90.15 - 90 = 0.15 \:minΔT=90.15−90=0.15min
So final answer :
\boxed{\sf{\huge{\red{\bold{\Delta t = 0.15 min}}}}}
Δt=0.15min