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A satellite is launched in a direction parallel to the surface of the earth with a velocity of $36900km/hr$ from an altitude of $500km$. Determine

(a) the maximum altitude reached by the satellite.

(b) the maximum allowable error in the direction of launching if the satellite is to go into orbit and come no closer than $200km$ to the surface of the earth.

I know that I have to apply the principle of conservation of angular momentum

$r_1 m v_1 \sin \phi_1 = r_2 m v_2 \sin \phi_2$

where $r =$ distance of the satellite from the center of the earth $m =$ mass of the satellite $v =$ velocity of the satellite $\phi =$ the angle the velocity makes with the radius vector

and the principle of conservation of energy

$T_1 + V_1 = T_2 + V_2$

Where $T =$ kinetic energy and $V =$ potential energy.

I know I have to use the two principles but I don't know how to use them. May someone please help me with how to use them or offer an alternative solution to the question.

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  • $\begingroup$ Write $T$ in terms of speed and $V$ in terms of position, and use the virial theorem. $\endgroup$
    – J.G.
    Mar 5, 2021 at 20:37
  • $\begingroup$ Am not so good with the virial theorem may be I'll have to dig into it $\endgroup$ Mar 5, 2021 at 20:45
  • $\begingroup$ Numerical computation gives the max height as 60299 km above ground (at 9.75 h) for the horizontal start. For an angled start the a minimal height of 200km above ground (at 19.42 h, max 60598 km at 9.67 h) is reached for 0.1996 rad = 11.436 deg. $\endgroup$ Mar 5, 2021 at 21:59
  • $\begingroup$ Seriously Lutz I didn't get you I am so sorry $\endgroup$ Mar 6, 2021 at 10:53
  • $\begingroup$ You can put the differential equation into the numerical solver of your choice, best with some event handling, or roll your own with RK4 or whatever is available, and then determine apogee and perigee from the numerical trajectory. A little tabulation and (manual) bisection rapidly finds the solution to the second task with 2 or 3 digits accuracy. This is not rocket science, or well, barely so. $\endgroup$ Mar 17, 2021 at 14:27

2 Answers 2

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I figured out part (a)

The altitude of the satellite is maximum and minimum for $\phi = 90^\circ$ because the satellite is revolving around the planet

thus from the conservation of angular momentum

$r_1 v_1 = r_2 v_2$

$r_2 = \frac{r_1 v_1}{v_2}$ --- (i)

Also from the conservation of energy

$T_1 + V_1 = T_2 + V_2$ --- (ii)

Where $T = \frac{1}{2}m v^2$ and $V = \frac{GMm}{r^2}$ where $G$ is the gravitational constant and $M$ is the mass of the earth

but $GM = g R^2$ where $R$ is the radius of the earth

So combining equation (i) and (ii)

$r_2 = \frac{v_1^2 r_1^2}{2g R^2 - v_1^2 r_1}$

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The Newton laws of gravity, on the way to establishing the Kepler laws, give a parametrization $$ r=\frac{R}{1+e\cos(φ-ψ)} ~~ \text{ and } ~~ \dot r=\sqrt{\frac{GM_E}{R}}e\sin(φ-ψ),~~r^2\dot\phi = -\sqrt{RGM_E} . $$ for the polar coordinates $(r(t),\phi(t))$, $x+iy=re^{i\phi}$, of the satellite (moving clock-wise if moving horizontally to the right). $ψ$ is the angle for the smallest radius, at periapsis, $\frac{R}{1+e}$.

The initial data in polar coordinates can be obtained from the given initial data that can easily be applied to Cartesian coordinates. As $x_0+iy_0=r_0e^{iφ_0}$, and the interpretation of the launch conditions as $x_0=0$, $y_0=R_E+H_{sat}$ gives $r_0=y_0$ and $φ_0=0$. The constants in the reduced equations follow from $$ \frac{\dot x_0+i\dot y_0}{x_0+iy_0}=\frac{\dot r_0}{r_0}+i\dotφ_0 $$ with $\dot x_0=V_{sat}\cos\theta$, $\dot y_0=V_{sat}\sin\theta$, so that $$ \dot r_0=\dot y_0=V_{sat}\sin\theta, ~~~ \dot φ_0=-\frac{\dot x_0}{y_0}=-\frac{V_{sat}}{r_0}\cos\theta $$ Then $$ r_0V_{sat}\cosθ=\sqrt{RGM_E}\implies R=\frac{r_0^2V_{sat}^2\cos^2θ}{GM_E}\\ e\cosψ=\frac{R}{r_0}-1 ~\text{ and }~ e\sinψ=-\dot r_0\sqrt{\frac{R}{GM_E}} $$ This is sufficient to calculate $e$ and $ψ$ as polar coordinates and thus the minimal and maximal elevations, periapsis and apoapsis (or perigee and apogee) $$ \frac{R}{1+e}-R_E ~\text{ and }~ \frac{R}{1-e}-R_E $$

G       = 6.67408e-11*1e-9 # km^3 s^-2 kg^-1
M_E  = 5.9721986e24 # kg
R_E  = 6378.137 # km
H_sat    = 500 # km above Earth
V_sat    = 36900/3600 # km/s
theta = 0

r0 = R_E + H_sat
dotr0 = V_sat*m.sin(theta)
dotphi0 = -V_sat/r0*m.cos(theta)

R = (r0**2*dotphi0)**2/(G*M_E)
wx = R/r0-1; wy = -dotr0*(R/(G*M_E))**0.5
E = (wx*wx+wy*wy)**0.5; psi = m.atan2(wy,wx)
print(f"R={R:.8f} km, e={E:.8f}, psi={psi:.8f} rad")
print(f"H_peri={R/(1+E)-R_E:.8f} km, H_apo={R/(1-E)-R_E:.8f} km")

with the result

R=12469.92166223 km, e=0.81297954, psi=-0.00000000 rad
H_peri=500.00000000 km, H_apo=60298.64124618 km

Setting theta=0.2 gives

R=11977.74005179 km, e=0.82116894, psi=-0.44435938 rad
H_peri=198.81463324 km, H_apo=60599.82661294 km

which is slightly outside the admissible range.

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