The difference in mechanical energy of a satellite

[Kare Skinnebach]

Homework Statement

This is from an exam for my current course in "Mechanics 2", from a couple of years ago, which means that the correct answers are given for checking results, but without further explanations. I'm confused about part (c) where my results do not correspond with the given answer, and I don't understand why.

My assignment is in Danish, so it's not verbatim, but I translated it as well as I could:
"To study an area with high solar activity at the Sun's equator, a satellite is sent into a heliostationary orbit (a satellite in a heliostationary orbit stays above the same point on the Sun's equator.

a) Find the distance from the center of the Sun at which the satellite will need to orbit, given that the Sun at the equator makes one rotation around its own axis in 27 days, 6 hours and 36 minutes. [solved]

b) The satellite starts at a point on the imaginary line going from the center of the Earth to the center of the Sun, where the gravity from the Sun and the gravity from the Earth cancel out. Calculate the distance from the center of the Earth to this point. [solved]

c) Determine the difference in the mechanical energy from the starting point to the heliostationary orbit for a satellite with a mass of 120 kg, given that the initial velocity at the starting point is $v_i=0$ m/s." [having trouble with this one]

We are given the following information:
Gravitational constant $G=6.67*10^{-11}$ Nm^2/kg^2
Mass of the Sun $M_s=1.99*10^{30}$ kg
Mass of the Earth $M_e=5.98*10^{24}$ kg
Radius of the Earth's orbit $r_e=1.50*10^{11}$ m

Homework Equations

1) Force of gravity $F_g=\frac{Gm_1m_2}{r^2}$
2) Potential energy due to gravity $U=-\frac{Gm_1m_2}{r}$
3) Period in circular orbit $2 \pi r*\sqrt{\frac{r}{G*m_e}}$

The Attempt at a Solution

To solve part (a), I used equation (3) above and set the period for the orbit equal to the rotation time of the Sun (converted to seconds), solve for $r$. $r=2.65*10^{10}$ m

(b) I made use of the fact that the force of gravity exerted on the satellite by the Sun and by the Earth would be equal at this starting position: $F_{gs}=F_{ge}$
I then used equation (2) above: $\frac{GM_e}{r_s}=\frac{GM_s}{(r_e-r_s)^2} \Rightarrow r_s=2.60*10^8$ m

(c) I figured that to calculate this I would need to find the difference in the sum of potential energy between the starting point and the heliostationary orbit. Is this wrong? Do I need to include the kinetic energy as well, and if so, how do I find it?
What I got was $\Delta E=(U_{sun,final}+U_{earth,final})-(U_{sun,initial}+U_{earth,inital})$ (total final potential energy minus total initial potential energy). This gives me $\Delta E = 4.938*10^{11}$ J, but the answer given is $1.9345*10^{11}$ J.

Help would be greatly appreciated. :)

 

[PeroK]

As your solution to a), you have $r = 2.65 \times 10^5m$

Isn't that inside the sun?

 

[mfb]

(b) is odd as well. While it might be the right answer to the problem statement, the problem ignores that Earth is rotating around the sun. The moon has a larger distance to Earth, and it still orbits Earth.

(c) I figured that to calculate this I would need to find the difference in the sum of potential energy between the starting point and the heliostationary orbit. Is this wrong? Do I need to include the kinetic energy as well, and if so, how do I find it?

You need the kinetic energy as well. You know the period and the radius of the satellite in the final orbit, how can you get the kinetic energy?A bonus question to get some sense of scale: How many tons of explosives (e. g. TNT) do you need to get this energy? Compare this to the 120 kg of the space probe.

 

[Kare Skinnebach]

As your solution to a), you have $r = 2.65 \times 10^5m$

Isn't that inside the sun?

Whoops, that was a typo. It's supposed to be $2.65*10^{10}$. And that's the distance from the center of the Earth.

 

[Kare Skinnebach]

(b) is odd as well. While it might be the right answer to the problem statement, the problem ignores that Earth is rotating around the sun. The moon has a larger distance to Earth, and it still orbits Earth.You need the kinetic energy as well. You know the period and the radius of the satellite in the final orbit, how can you get the kinetic energy?A bonus question to get some sense of scale: How many tons of explosives (e. g. TNT) do you need to get this energy? Compare this to the 120 kg of the space probe.

Yeah that is kind of weird.
For the kinetic energy, I think I just found the right formula: I found v from $T=\frac{2\pi r}{v}$, then used $K=\frac{1}{2} mv^2$ to find kinetic energy and just added that to the potential energy difference, is that right?
I get $K=1.9364*10^{11}$, but the correct answer is $1.9345*10^{11}$, but that could just be a rounding error or something.
By my calculations you would need $46.28$ tons of TNT, which is quite a lot compared to the weight of the satellite, hadn't really thought much about that number. Thank you for that, haha! :)

 

[mfb]

The small difference should be a rounding error, right.