A weather satellite ($m_s = 4350$ kg) is in a stable circular orbit around the Earth ($m_E = 5.97 \cdot 10^{24}$ kg). It completes an orbit once every 2 and a half hours.

(I'm sure about these 2 answers) At what distance from the center of the Earth does the satellite orbit? $r_i = 9.35 \cdot 10^6$ m

What is the angular velocity of the satellite? $\omega_i = 6.98 \cdot 10^{-4}$ rad/s

(I'm having trouble getting this part:) The satellite operators decide to fire maneuvering rockets and move the satellite into an orbit with a 5% larger radius. If the initial magnitude of the satellite’s mechanical energy was $E_{m,i} = 9.26 \cdot 10^{10}$ J and it continues at the same speed, how much work was done by the rockets in moving the satellite to the higher orbit?

I calculated the larger radius to be $r_f= 9.86 \cdot 10^6$ m.

Using $v = \sqrt{\frac{GM_e}{r}}$, I found $v_i = \sqrt{\frac{GM_e}{r_i}} = 6529.7$ m/s. Likewise, $v_f=6371.5$ m/s.

Using the work-energy theorem, I know that $W_i+K_i+W_{other}=W_f+K_f$. Because the total mechanical energy is given, $E_{m,i}+W_{other}=W_f+K_f$. The only work done on the object is by potential energy due to gravity, so $E_{m,i}+W_{other}=-\frac{GM_em_s}{r_f}+\frac{1}{2}m_sv_f^2 \implies W_{other}=-\frac{GM_em_s}{r_f}+\frac{1}{2}m_sv_f^2-E_{m,i} = -1.80 \cdot 10^{11}$ J, which is clearly the wrong answer. Can somebody spot where I'm messing up?

  • $\begingroup$ Welcome to Physics! Generally we discourage questions that just ask for someone to check your work. Can you try to identify the specific concept(s) that you're not sure about, which might have caused you to make a mistake, and ask about that? $\endgroup$
    – David Z
    Apr 12, 2012 at 23:55
  • $\begingroup$ @DavidZaslavsky, sorry about not asking a specific question; I'm having trouble grasping this conceptually and I thought it would be better to just put what I had rather than nothing. I'll keep what said in mind for the future, though! $\endgroup$ Apr 13, 2012 at 0:03
  • $\begingroup$ on the left hand side of your energy equation, you forgot to include the gravitational potential at the initial radius. $\endgroup$
    – user1631
    Apr 13, 2012 at 0:13
  • $\begingroup$ @user1631, why isn't that included in the total mechanical energy that's given? $\endgroup$ Apr 13, 2012 at 0:55
  • $\begingroup$ Apparently by "mechanical" they mean "kinetic". Total kinetic + gravitational will be negative. $\endgroup$
    – user1631
    Apr 13, 2012 at 1:31

3 Answers 3


I'm going to offer a completely different way of doing this. Sometimes it's nice to work in symbols before getting into the very specific numbers. I take my inspiration from the Bohr atomic model here. The total energy of the electron in orbit around the nucleus at any given radius is calculated as follows (Wikipedia's equation here, not mine).

Bohr energy level

That is, the sum of the kinetic and potential energy is just half the potential energy! Neat, isn't it? So, how would we apply this to the Earth? Well, $Z k_e e^2$ is going to have to be replaced with $G M m$. But let's not forget, the entire point was to introduce a radius $r'=1.05 r$, and I'm seeking a value of $\Delta E = E'-E$. Also, the kinetic energy is 1/2 the magnitude of this total energy metric, I'll use $E_k$ for that.

$$\Delta E = GMm/2 \left( -1/r' + 1/r \right) = E \left(-1/1.05+1\right)=E\frac{0.05}{1.05} = 2 E_k \frac{0.05}{1.05}$$

So the energy would change by about 9.5% times the original kinetic energy. Given your original energy, I believe this would be $8.8 \times 10^9 J$.

This all said, your question says:

If the initial magnitude of the satellite’s mechanical energy was $E_{m,i} = 9.26 \cdot 10^{10}$ J and it continues at the same speed, how much work was done by the rockets in moving the satellite to the higher orbit?

Our work assumed that it would attain a new speed. Maybe the question is written wrong. I don't know.

  • $\begingroup$ Oh drat, the energy given in the question is the original kinetic energy. So I think that is 3 times the value that I needed. Divide my answer by 3. $\endgroup$ Apr 13, 2012 at 0:52
  • $\begingroup$ Hm, $8.82 \cdot 10^9$ J is the right answer. $\endgroup$ Apr 13, 2012 at 0:56
  • $\begingroup$ Well, I keep messing up these factors. It's a factor of 2 it was off by I think. I also randomly replace a 9 with a 6, all combined I think I might come to @user28554's number. $\endgroup$ Apr 13, 2012 at 0:59
  • $\begingroup$ So, where does this factor of 2 come in at $E\frac{0.05}{1.05}$? Interesting approach, by the way! $\endgroup$ Apr 13, 2012 at 1:04
  • $\begingroup$ @user28554 The novelty of the Bohr atom approach is the observation that for circular orbits the gravitational potential energy is two times the kinetic energy. Of course, this is complicated by the fact that the gravitational potential energy is negative, since it is referenced to r=infinity. The 3 was me making the dumb mistake of switching 1-1/2 for 1+1/2. $\endgroup$ Apr 13, 2012 at 1:18

By using equations from your very post, it is clear that the total mechanical energy is

$$E = U + K = -\frac{GMm}{r}+ \frac{1}{2} \frac{GMm}{r} = - \frac{1}{2} \frac{GMm}{r}.$$

So total mechanical energy increases with larger orbit. Work done is difference between final and initial mechanical energy

$$W = E_f - E_i$$

which is thus positive.


You need to take potential and kinetic energy into account for an object in a gravitational field.

Potential: $E_p = - mgh$

Kinetic: $E_k = \frac{1}{2}mv^2$

Energy: $E = E_{k} + E_{p} = \frac{1}{2}mv^2 - mgh$

For a circular orbit, $h = r$

$g = \frac{GM}{r^2} = \frac{\alpha}{r^2} \rightarrow E_p = - m(\frac{GM}{r^2})r= - \frac{m\alpha}{r}$

and for circular motion $a = g = \frac{v^2}{r} \rightarrow v^2 = gr = \frac{\alpha}{r} \rightarrow \frac{1}{2}mv^2 = \frac{m\alpha}{2r}$

Energy (Circular orbit, a=g): $E = \frac{m\alpha}{2r} - \frac{m\alpha}{r} = - \frac{1}{2}\frac{m\alpha}{r}$

Difference: $\Delta E = E_2 - E_1$

So: $\Delta E = - \frac{1}{2}m\alpha (\frac{1}{r_2} - \frac{1}{r_1})$

For $r_2 = 1.05\ r_1$ we get $\Delta E = \frac{1}{2}GMm(\frac{0.05}{1.05\ r_1})$

I've assumed you meant the same angular velocity, because that would make more sense in this case, since the velocity of the satellite in free orbit is determined uniquely by the radius of its orbit.

  • 1
    $\begingroup$ Ep=mgh? That's not going to work here. $\endgroup$
    – tmac
    Apr 13, 2012 at 0:44
  • $\begingroup$ Hmm... I suppose a $g_{eff}$ would be more appropriate... $\endgroup$ Apr 13, 2012 at 0:57
  • $\begingroup$ Your "derivation" of potential energy by simply multiplying with $r$ is very problematic. In fact, derivation of potential energy requires integration of force with changing radius $U = \int_\infty^r -\left(-\frac{GMm}{r^2}\right)$. $\endgroup$
    – Pygmalion
    Apr 13, 2012 at 6:05
  • $\begingroup$ I figured an integral here would be too much detail for the required answer. And it doesn't really matter for such a simple potential. $\endgroup$ Apr 16, 2012 at 5:53

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