Spaceship Orbit

As a motion in a curved path involves changing of direction of the velocity vector, it is an accelerated motion. It requires a force toward the center of curvature called centripetal force given by $F_c = \dfrac {mv^2}r$ , where $m$ is the mass of the object in motion, $v$ , the velocity of the object and $r$ , the radius of curvature.

If $F_c > mg$ , where $g$ is the acceleration due to gravity, then the object will spin off the curve. If $F_c < mg$ , the object will move towards Earth. For the object to stay in orbit, then $F_c = mg$ . Then $\dfrac {mv^2}r = mg$ , $\implies v = \sqrt{rg} = \sqrt{(400+6371)10^3\times 8.69} \approx \boxed{7670.723}$ .

The spaceship will remain in orbit so long as the magnitudes of the centripetal and gravitational forces acting on it balance, that is $F_c = F_g\\ \frac{m_sv^2}{r} = \frac{Gm_sm_e}{r^2}$ where $m_s$ is the mass of the spaceship and $m_e$ is the mass of the earth $m_e = 5.972\cdot 10^{24}kg$ , and $G$ is the gravitational constant $G=6.67\cdot10^{-11}\frac{Nm^2}{kg^2}$ . Solving for velocity gives $v = \sqrt{\frac{Gm_e}{r}}, \, \,\,\, \text{where } r = (6371+400)\cdot10^3 m$ $v= \sqrt{\frac{6.67\cdot10^{-11}\cdot5.972\cdot 10^{24}}{ (6371+400)\cdot10^3}} \approx \boxed{7670}.$

In order to let the spaceship at 400 km height (above ground) go around the earth without spiraling down/flying away from it, the centrifugal force on the spaceship has to be equivalent to / delivered by the gravitational force. So:

$F_c = F_g$

$\frac{mv^2}{r+h}= m \cdot \frac{GM}{(r+h)^2}$ , where $m$ is the mass of the spaceship (kg), $M$ is the mass of the earth (kg), $v$ is the velocity of the spaceship (m/s), $G$ is the gravitational constant $(6.67 \cdot 10^{-11} Nm^{2}kg^{-2})$ , $r$ is the radius of the earth (m) and $h$ is heigth of the spaceship above the ground (m).

$M$ is not given, but we still can use that $a=8.69$ m/s $^2$ , because $\frac{GM}{(r+h)^2}=a$ .

Plugging this last formula in the original equation will result into:

$\frac{mv^2}{r+h}= m \cdot a$ , where $a$ is the acceleration which is pulling the spaceship down were it not moving forward (m/s $^2$ ).

Multiplying both sides by $(r+h)/m$ , will give us $v^2=a(r+h)$ , which means $v=+\sqrt{a(r+h)}=\sqrt{8.69((6371+400)\cdot 10^3)}\approx \boxed{7.67\cdot10^3}$ (m/s).

Solution with no physics formulas only basic calculus:

Let $t$ be the number of seconds after the rocket was directly right of Earth (0 degrees). Let $x(t)$ and $y(t)$ be the be rocket's position relative to the center of the Earth in meters. The given details show us that $x(0)=6771000$ and $x''(0)=-8.69$ . Let $z$ be the number of seconds it takes the rocket to orbit the earth.

$x(t) = 6771000 \cos(\frac{2 \pi}{z}t)$

$x''(t) = -6771000 (\frac{2 \pi}{z})^2 \cos(\frac{2 \pi}{z}t)$

Plug 0 into $x''(t)$ :

$-8.69 = -6771000 (\frac{2 \pi}{z})^2$

$\frac{6771000}{8.69} = (\frac{2 \pi}{z})^2$

$z=2 \pi \sqrt{\frac{6771000}{8.69}}$

The distance the rocket travels is given by $2 \pi r$ where r is the distance to the center of the earth. Plugging in r we get: $13542000 \pi$ Now that we have the distance in meters and time in seconds it takes to orbit the earth, we can just take $\frac{distance}{time}$ to find the velocity.

$\frac{2 \pi \sqrt{\frac{6771000}{8.69}}}{13542000 \pi}=\boxed{7670.723}$

I'm not familiar with orbital equations and I don't know the formula for centripetal force, but I have a basic understanding of physics and math, so my solution just uses the Pythagorean theorem, limits, and an equation of motion.

The ship is distance $r$ from the center of the earth.

The ship moves forward some distance $d$ over a time period $\Delta t$

Using this equation of motion $x = x_o + vt + \frac{1}{2}at^2$ we find the ship then falls towards the earth for a distance $\frac{1}{2}g\Delta t^2$ and since the ship is orbiting, the remaining distance to the center is $r$

This forms a right triangle, thus we can use the Pythagorean theorem to find $d^2+r^2=(\frac{1}{2}g\Delta t^2+r)^2\implies d=\sqrt{(\frac{1}{2}g\Delta t^2+r)^2-r^2}$

Now, in reality the ship falls toward the earth at the same time that it is moving, so there is some error. But the error gets smaller as $\Delta t$ gets smaller.

So we can find the speed using the equation $S = \lim_{\Delta t \to 0}d/\Delta t= \lim_{\Delta t \to 0}\sqrt{(\frac{1}{2}g\Delta t^2+r)^2-r^2}/\Delta t$

$\implies S = \lim_{\Delta t \to 0}\sqrt{\frac{1}{4}g^2\Delta t^4+rg\Delta t^2+r^2-r^2}/\Delta t$

$\implies S = \lim_{\Delta t \to 0}\sqrt{\frac{1}{4}g^2\Delta t^4+rg\Delta t^2}/\Delta t$

$\implies S = \lim_{\Delta t \to 0}\Delta t\sqrt{\frac{1}{4}g^2\Delta t^2+rg}/\Delta t$

$\implies S = \lim_{\Delta t \to 0}\sqrt{\frac{1}{4}g^2\Delta t^2+rg}$

$\implies S = \sqrt{rg}$

Plugging in the values $r = \SI{400000}{\meter}+ \SI{6731000}{\meter}$ and $g=\SI{8.69}{\meter/\second^2}$ give the answer $S=\SI{7670}{\meter/\second}$

g = $v^2/R$ where g = 8.69 $m/s^2$ and R =400000+6371000 m

thus $v = sqrt(mR)$

so $v = sqrt(8.68 \times 6771000)$ =7671.7229

It is worth pointing out that this is the required speed for a circular orbit. There is a range of lower speeds that would result in an elliptical orbit with an apogee of 6771 km.

Also, too slow a speed would not result in a spiraling descent, merely an ellipse whose perigee would be less than 6371 km, resulting in a bad day for the astronauts.

in a circular trajectory there is always an acceleration which direction is the center, even if the motion has constant velocity. This acceleration has a value of:

$a=\frac{v^2}{r}$

$a:$ acceleration.

$v:$ velocity.

$r:$ radius of the circular trajectory

As we can imagine the only acceleration which keeps the rocket orbiting the planet in a constant circular trajectory is the gravity, so we can construct the next equation:

$g=8.69=\frac{v^2}{r}$

now we need the radius of the circular trajectory to get the velocity we need. We obtain this radius adding the distance between the earth surface and the rocket, with the earth radius, as following:

$r=400 [km]+6371 [km]=6771 [km]=6771 [km]*\frac{1000 [m]}{1 [km]}=6771000 [m]$

We transform it into meters because we are asked to give the result in $\frac{m}{s}$ , and we have to be congruent with the units. Now we can find the velocity with the equation we have constructed:

$v=\sqrt{8.69 [\frac{m}{s^2}]* 6771000 [m]}=7670.722912[\frac{m}{s}]$

Actually, there is a flaw in the question. If the spaceship is orbiting at a fixed distance from the centre of the earth, gravitational acceleration " ON THE SPACESHIP " would be zero(experiencing free fall). Instead, it should have been given that the gravitational acceleration is 8.69m/s2 at that height.

Instead of using the fact that the force of gravity and the centripetal force have to be equal. I used the equation for centripetal acceleration $a_c = \dfrac{v^2}{r}$ . Then I solved for $v$ to get the equation $v = \sqrt{ra_c}$ . Then, I plugged in the value for the centripetal acceleration (which is the gravitational acceleration in this case). And for the radius of the rotation. Which is the radius of the orbit plus the radius of the earth; because the radius we put into our formula has to be the same as the radius of orbit if the earth was just a point with the same mass as the earth. Plug the values in and one gets $7670.72\cdots$