Finding r of a Jovian-synchronous orbit and escape velocity

In summary, the satellite would need to remain 1.6*10^8 m above the equator of Jupiter for a jovian-synchronous orbit, and the escape velocity from Jupiter is approximately 60,000 ms^-1. However, the wording of the question for part 1 is unclear about the meaning of r, but your calculations for the escape velocity in part 2 are correct.
  • #1
AN630078
242
25
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
 
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  • #2
Looks good for both parts. But for part 1 you concluded that
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.
Does r represent the distance above the surface of Jupiter or the distance from the center of Jupiter?
 
  • #3
TSny said:
Looks good for both parts. But for part 1 you concluded that
Does r represent the distance above the surface of Jupiter or the distance from the center of Jupiter?
Thank you for your reply. I was using r to represent the distance above the surface of Jupiter I believe, woudl this be wrong?
 
  • #4
AN630078 said:
Thank you for your reply. I was using r to represent the distance above the surface of Jupiter I believe, woudl this be wrong?
r denotes the radius of the circle that represents the orbit of the satellite. It's not really clear to me what the question means by "distance from Jupiter".
 
  • #5
TSny said:
r denotes the radius of the circle that represents the orbit of the satellite. It's not really clear to me what the question means by "distance from Jupiter".
I think the "distance from Jupiter" is supposed to be the distance from the equator at the surface of Jupiter to the height of the orbiting satellite. Would be workings be incorrect then?
 
  • #6
AN630078 said:
I think the "distance from Jupiter" is supposed to be the distance from the equator at the surface of Jupiter to the height of the orbiting satellite. Would be workings be incorrect then?
I believe that you got the correct answer for r. But r is not the distance between the surface of Jupiter and the satellite.
 
  • #7
TSny said:
I believe that you got the correct answer for r. But r is not the distance between the surface of Jupiter and the satellite.
Thank you for your reply. Oh ok, should I try to amend this or is it the wording of the question that is erroneous? Also, would my calculations for the escape velocity be correct?
 
  • #8
AN630078 said:
Thank you for your reply. Oh ok, should I try to amend this or is it the wording of the question that is erroneous? Also, would my calculations for the escape velocity be correct?
For me, the problem statement for part 1 is not too clear. The fact that they don't give you the radius of Jupiter until part 2 makes me think that they are just asking for r in part 1, rather than asking for the height above the surface of Jupiter. But, I'm not sure. Since this is just a practice problem, I wouldn't worry about it as long as you understand the meaning of r and how you could use r and the radius of the planet to calculate the height above the surface.

Your work for part 2 looks good to me.
 

1. What is a Jovian-synchronous orbit?

A Jovian-synchronous orbit is an orbit around a Jovian (Jupiter-like) planet that has a period of exactly one day, meaning it completes one orbit in the same amount of time it takes the planet to rotate on its axis.

2. How do you calculate the radius of a Jovian-synchronous orbit?

The radius of a Jovian-synchronous orbit can be calculated using the formula r = (GMpT^2/4π^2)^(1/3), where G is the gravitational constant, Mp is the mass of the planet, and T is the planet's rotational period.

3. What is the escape velocity of a Jovian-synchronous orbit?

The escape velocity of a Jovian-synchronous orbit is the minimum speed required for an object to escape the gravitational pull of the planet and enter into an unbound orbit. It can be calculated using the formula Ve = √(2GMp/r), where G is the gravitational constant, Mp is the mass of the planet, and r is the distance from the center of the planet to the object.

4. How does the mass of the planet affect the radius of a Jovian-synchronous orbit?

The mass of the planet has a direct effect on the radius of a Jovian-synchronous orbit. The greater the mass of the planet, the larger the radius of the orbit will be. This is because the stronger gravitational pull of a more massive planet requires a larger distance for an object to maintain a synchronous orbit.

5. Can a spacecraft in a Jovian-synchronous orbit reach escape velocity?

Yes, a spacecraft in a Jovian-synchronous orbit can reach escape velocity by increasing its speed to the required velocity. However, this would require a significant amount of energy and is not a practical option for most spacecraft missions. It is more common for spacecraft to use a gravity assist from the planet to achieve escape velocity.

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