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Homework question

I have attached an image above including the question I'm unsure of (part B)

In a nutshell, a satellite is orbiting Earth at the surface. Given is $g$ and the radius of the Earth.

For part bi) the question is to calculate the speed of the satellite; here is my working: working to part bi

I am confused about why we take the centripetal acceleration to be $9.8~\mathrm{ms^{-2}}$. Surely the satellite is able to travel faster or slower than the calculated $7900~\mathrm{ms^{-1}}$, and as a result the centripetal acceleration would be correspondingly higher or lower than $9.8~\mathrm{ms^{-2}}$ (following the equation $a=\frac{v^2}{r}$).

I understand that the centripetal force is being provided only by gravity, however I do not understand why a must be 9.8.

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  • $\begingroup$ There is only one velocity that will give rise to a circular orbit at this distance. Slower and it will crash; faster, and it will become elliptical. $\endgroup$
    – Floris
    Commented Jul 25, 2016 at 18:32
  • $\begingroup$ Thanks for the reply! Is this the same in any system involving gravity, for example with the Earth orbiting the Sun, there is only one velocity which enables it to orbit at a given distance? $\endgroup$
    – John
    Commented Jul 25, 2016 at 18:42
  • $\begingroup$ Yes - if you are looking for a stable circular orbit, there is exactly one velocity that will keep you there. This is fundamental. $\endgroup$
    – Floris
    Commented Jul 25, 2016 at 18:43
  • $\begingroup$ But for cases where gravity is not the centripetal force, for example a rotating turntable, there is not just one velocity for a stable circular motion? $\endgroup$
    – John
    Commented Jul 25, 2016 at 19:05
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    $\begingroup$ Now it makes sense. Thank you so much for all your help! $\endgroup$
    – John
    Commented Jul 25, 2016 at 21:11

2 Answers 2

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Assuming no air resistance, the only force acting on the satellite is the force of gravity, $\bf{F_g}$.
The stated circular orbit also means that centripetal force, $\bf{F_c}$ is involved, which causes the satellite to follow the circular path.
Due to this, the centripetal acceleration can be directly equated to g, which allows you to immediately solve for velocity.

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You are orbiting at the surface of the earth, so $r$ is fixed. Since $r$ is fixed you are not free to change $a$; it is fixed by Newton's Law of gravitation $$ a = \frac{F}{m} = G\frac{m_\mathrm{earth}}{r_\mathrm{earth}^2} = 9.8\;\;\mathrm{m/s}^2$$ Since $a$ is fixed and $r$ is fixed, $v$ is determined.

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  • $\begingroup$ Thanks for the reply! Even though r is fixed, is it not possible for the satellite to increase its speed, for example by using rockets attached to it, thereby increasing the acceleration and the velocity? $\endgroup$
    – John
    Commented Jul 25, 2016 at 18:36
  • $\begingroup$ Not if the orbit is to remain circular with a radius equal to the earth's radius. You can increase the speed with rockets and put the satellite into an elliptical orbit that just touches the earth's surface twice each orbit. You can put a satellite into orbit at any radius you want at any speed if you have a rocket that fires continuously ... until you run out of fuel. $\endgroup$
    – garyp
    Commented Jul 25, 2016 at 18:43
  • $\begingroup$ when you say you can put it into orbit at any radius at any speed, do you mean only one speed for a given radius? Or by "orbit" were you including elliptical orbits? (Sorry to be a bother) $\endgroup$
    – John
    Commented Jul 25, 2016 at 18:55
  • $\begingroup$ @John , each radius has its own unique velocity when in a circular orbit. $\endgroup$
    – David White
    Commented Jul 25, 2016 at 22:40
  • $\begingroup$ Normally, when we talk about orbits, we are talking about objects revolving under the influence of gravity alone. In that case, the velocity is fixed for each radius. You mentioned rockets. A rocket engine that turns on then turns off will result in an elliptic orbit. It is possible to configure a rocket engine to burn continuously and arrange any velocity at any radius. We don't normally call that an orbit, though. And it would be sustainable only until the fuel runs out ... seconds, minutes, or perhaps hours. $\endgroup$
    – garyp
    Commented Jul 25, 2016 at 23:43

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