Question

A satellite of mass m (where m ≪ Me) is initially in a circular orbit around the Earth at a height of 410 km above the Earth’s equator. Its operators would like to move it into a geosynchronous orbit using a Hohmann transfer orbit. Assume a spherical Earth with radius 6371 km.

(a) Sketch the satellite’s Hohmann transfer orbit.

(b) Find the satellite’s initial (circular) orbital speed according to an inertial observer.

(c) Find the maximum height of the satellite above the Earth’s surface when it is on its Hohmann transfer orbit.

(d) How much does the satellite’s initial speed need to be increased to place it on its Hohmann transfer orbit?

Answer 1

PART A

earth AV

the initial speed can be calvulated with the hep of the formula

Vo = GM V 91

Vo = orbital velocity

G = universal gravitational constant = 6.67 x 10 -11 Nm2kg-2

M = mass of earth (standarded value approximate 6.0 x 10 24 kg)

r = radius of orbit from the center on the earth

for the question it is given that the

radius of earth 6371 km

height or orbit from earth surface = 410 km

so radius of orbit r = 6371+410 = 6781 km

for part B initial orbital velocity Voi =

6.67 X10 X 6x1024 6781 Y1000

Voi = 7682.30 m/s

part c the maximum height = ?

maximum height is equal to the height or radius of geosynchronous orbit

for identifing the maximum height

as per the defination of geosynchonous orbit the angular velocity of geosynchronous is equal to the angular velocity of earth on its axis.

the angular velocity of eath is equal to angular velocity = 2π / T

T = 24 hrs

angular velocity of earth = 7.3 x 10-5 rad/sec

so the velocity of geosynchronous Vgy = radius x angular velocity(ω)

velocity formula of a orbit =

Vo = GM V 91

Vgy =

                          

Vag hw - LG hwe = GM

by putting all value of G, M and angular velocity we will get =42549 km (approx)

so the maximum height of the satallite from the surface of earth is = 42549-6371= 36178 km

for part D we have to known the preigee and apogee

as the shown in figure we gave to calculate the velocity at pregee and apogee

Vp = velocity at the perigee Va = velocity at the apogee M = Earth >Va P = radius of orbit at the perigee A = radius of orbit at the apogee c= distance from center to one focus a = distance from center to one vertex (half the major axis)

We know that eccentricity is the ratio of cto a ore = -. So, all that is left to do is put P and A in terms of a and c and solve for a. Looking at our diagram we get, P = a - c p = a - ea p = a(1-e) A = a + c A=a +ea A = a1 + e) PA So, a= — and a 1-e 1 + e Therefore, vp-Scm (43")or simplified, Vo -. /ov (1)

42549-6781 eccentenci 42549 +678) C = 0.73 U periger - I GM Cire) GM Cite) Uperigel 6.67X10" x 6 x 10 x C1+0.13) 6781x10² Uperigee- Canon 10104.5 m/s same GM CI-Cl Vapogee = 1 u2sug adgee - 12620th me m/s 1593.6 Vapogee =

The satellite travels at a speed of 7682m/s (part A) around the small circle. We need to speed the satellite up to 10104.5m/s (pregee) when it reaches the perigee so it will transfer onto the ellipse. We then need to slow the satellite down to 1593.6m/s at the apogee, and then speed it up again to 3100m/s (Vgy) to transfer it to the higher circle that orbits at 42549 km above the Earth’s surface.