Question

2. Consider a system consisting of the Sun, Earth, and a satellite in a circular orbit about the Earth. (a) Plot the gravitational acceleration of the satellite due to Earth's gravity as a function of the altitude of the satellite as measured from the surface of the Earth. Scale your plot so that the altitude goes from 0 km to 104 km. (b) Assume that the satellite instantaneously lies on the line between the Earth and the Sun. Make two plots of the gravitational acceleration of the satellite due to Earth's gravity and of the gravitational acceleration of the satellite due to the Sun's gravity, each as a function of the altitude of the satellite as measured from the surface of the Earth. Scale your plot so that the satellite altitude goes from 0 km to 106 km. (C) Using your plots in part (b), assess the relative importance of Earth's gravity and Sun's gravity on the satellite's orbit. For what satellite altitudes do you think it is a reasonable approximation to ignore the Sun's gravity and use a two-body model for the satellite's orbit based only on Earth's gravity? (d) Using your part (b) plots again, for what altitudes do you think it is reasonable to ignore the Earth's gravity and use a two-body model for the satellite's orbit based only on the Sun's gravity? (NB. Below this altitude the satellite is said to be within the Earth's sphere of influence and above it the satellite has "escaped" from the Earth and is now moving under the influence of the Sun.

Answer 1

(a)The law of gravity states that the force of gravity is additive and the force between two bodies of mass m_1 and m_2 and separated by distance d, due to gravity will be,

Gmim2 F=

In our case, let us use the following abbreviations

=mass of satellite *this will become irrelevant once we go further)

$m_e$=mass of earth=6x10^24 kg

$m_s$=mass of the sun=2x10^30 kg

L= distance between earth and sun=1.5x10^8m

h=distance of the satellite from earths surface

$r_e$=radius of the earth=6.37x10^6m

G=universal gravitational constant=6.67x10^-11 Nm^2/kg^2

When the satellite is at a height of h from the surface of the earth, it is at a height of (h+r_e) from the center of the earth. So, the force of pull due to earths gravity will be

Gmme F h+re)'

if acceleration due to gravity is g(h) then

C Gme q(h) = - = 6.67 x 10-11 x 6 x 1024 (h+6.37 x 1072 4 x 1014 (h + 6.37 x 107) m (h+r)2

The data and graph of h and g(h) is shown below.

Tech) h km m/s^2 9.86 9.83 100 9.56 500 8.48 1000 7.36 2000 5.71 3000 4.56 4000 3.72 5000 3.09 2.61 7000 2.24 8000 1.94 9000 1.69 10000 1.49 6000

g(h)vsh 12.00 Acceleration due to gravity, m/s^2 2000 10000 12000 4000 6000 8000 Height from Earth's surface,km

(b)Continuing on the logic of the previous problem, when the satellite is at height h above the earth's surface, and is instantaneously on the line between the earth and the sun, it is (L-h-r_e) from the center of the sun. So, the total force on the satellite towards the sun will be

F = Gmms (L-h-re)?

and the acceleration will be

g(h), = Gm (L-h-re)?

The acceleration due to earths gravity will be (same as earlier arrived)

Gm g() = (1 +T)*

Substituting numbers,

g(h), = 1.33 X 1020 (1.5 x 1011 - h)

g(h)e = 4 x 1014 (h +6.37 x 107)

Plotting, we get:

km 0 g(h) e g(h) s m/s^2 m/s^2 9.86 0.0059 1000 7.36 0.0059 5000 3.09 0.0059 10000 1.49 0.0059 20000 0.58 0.0059 50000 0.13 0.0059 100000 0.04 0.0059 200000 0.01 0.0059 500000 0.00 0.0060 1000000 0.00 0.0060

g(h) vs h 12.00 10.00 g(h)_e, m/s^2 0.0060 0.0060 0.0060 0.0060 0.0060 0.0060 0.0059 60 0.0059 0.0059 0.0059 0.0059 gth_s gh)_s, m/s^2 Series1 0 200000 1000000 0.00 1200000 400000 500000 800000 Height from Earth's surface, km

(c) To find out more, let us plot $\frac{g(h)_s}{g(h)_e}$ as a function of h.

km g(h)e g(h) 5 g(h)_s/g(h) e m/s^2 m/s^2 op 9.85781 0.0059 0.001 1000 7.3642 0.0059 0.001 5000 3.0941 0.0059 0.00 10000 1.4927 0.0059 0.00 20000 0.57521 0.0059 0.01 50000 0.1259 0.0059 0.051 100000 0.0354 0.0059 180000 0.0115 0.0059 200000 0.0094 0.0059 0.63 500000 0.0016 0.0060 3.81 1000000 0.0004 0.0060 15.17

From the above table, we see that the earths gravity is dominant at 2x10^4 km, where the sun's gravity is 1% of the earth's gravity. At 1.6x10^5 m, the suns gravity starts to exceed earths gravity. Assuming 1% error to be acceptable in approximations, the sun's gravity can be ignored till 2x10^4 km.

(d) By observing the previous table, we see that at 1.8x10^5 km, the suns gravity greater than the earths gravity, and at this point, it can be said that the satellite has broken from the earths gravity and is under the major influence of the sun.