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AN630078
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- Homework Statement
- Hello, I have recently been introduced to the principle of escape velocity, in additional to other gravitational concepts, and finding the orbital period. I was attempting some practise questions I found online but I wondered whether anyone could comment upon my workings and calculations, particularly to suggest any improvements if I am faced with similar problems. I would be very grateful of any help
1.The mass of Jupiter is 1.89 x 10^27 kg. It rotates on its own axis in 9 hours and 55 minutes. At
what distance from Jupiter would a satellite above the equator remain always above the same spot?
2. The radius of Jupiter is 7.1 x 107. Calculate the escape velocity from Jupiter
- Relevant Equations
- T^2=4π^2r^3/GM
Ui + Ki=Uf+Kf
1. The satellite would be in a jovian-synchronous orbit,
Rearranging the formula for the orbital period in terms of r, since T^2 is proportional to r^3:
T^2=4π^2r^3/GM which becomes r^3=(GM/4π^2) T^2
M(mass of Jupiter)=1.89 x 10^27
G=6.67*10^-11 m^3kg^-1s^-2
T=9 hours and 55 minutes = 3.57*10^4 s
Therefore, r^3= (6.67*10^-11*1.89 x 10^27 /4π^2) *(3.57*10^4)^2
r^3=4.069718155*10^24 m
Taking the cube root of both sides;
r=159657053.9 ~ 160,000,000 m or 1.6*10^8 m
So the satellite would need to remain 1.6*10^8 m above the equator of Jupiter.
2. The escape velocity is the minimum speed required for a free, non-propelled object to escape from the gravitational influence of a massive body, that is, to achieve an infinite distance from it.
Applying the conservation of energy;
Ui + Ki=Uf+Kf
Set both terms on the right to zero, since we want the object to reach infinity, where the potential energy is zero.
For a planet of mass M and radius R, the potential energy of an object of mass m at the planet's surface is: U= - GMm/R.
Therefore, - G Mm/R + K escape =0
1/2mv^2 escape = GMm/R
Cancel m and multiply both sides by 2:
v escape = √ 2GM/R
v escape =√ 2*6.67*10^-11*1.89 x 10^27 /7.1 x 10^7
v escape = 59590.85855 ~60,000 ms^-1
Rearranging the formula for the orbital period in terms of r, since T^2 is proportional to r^3:
T^2=4π^2r^3/GM which becomes r^3=(GM/4π^2) T^2
M(mass of Jupiter)=1.89 x 10^27
G=6.67*10^-11 m^3kg^-1s^-2
T=9 hours and 55 minutes = 3.57*10^4 s
Therefore, r^3= (6.67*10^-11*1.89 x 10^27 /4π^2) *(3.57*10^4)^2
r^3=4.069718155*10^24 m
Taking the cube root of both sides;
r=159657053.9 ~ 160,000,000 m or 1.6*10^8 m
So the satellite would need to remain 1.6*10^8 m above the equator of Jupiter.
2. The escape velocity is the minimum speed required for a free, non-propelled object to escape from the gravitational influence of a massive body, that is, to achieve an infinite distance from it.
Applying the conservation of energy;
Ui + Ki=Uf+Kf
Set both terms on the right to zero, since we want the object to reach infinity, where the potential energy is zero.
For a planet of mass M and radius R, the potential energy of an object of mass m at the planet's surface is: U= - GMm/R.
Therefore, - G Mm/R + K escape =0
1/2mv^2 escape = GMm/R
Cancel m and multiply both sides by 2:
v escape = √ 2GM/R
v escape =√ 2*6.67*10^-11*1.89 x 10^27 /7.1 x 10^7
v escape = 59590.85855 ~60,000 ms^-1