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algol [13]
3 years ago
8

Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6

.67 x 10-11 m3kg-1s-2, and LEO is 400 km above Earth's surface.
Physics
2 answers:
Tamiku [17]3 years ago
6 0

Answer:9.8m/s^2

Explanation:

Acceleration due to gravity(g)=?

Mass of earth(m)=6x10^24kg

G=6.67x10^(-11)

Radius of earth(r)=6.4x10^6m

g=(mxG)/r^2

g=(6x10^24x6.67x10^(-11))/(6.4x10^6)

g=(4.002x10^14)/(4.096x10^13)

g=9.8m/s^2

Nana76 [90]3 years ago
4 0

Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }

Where F_{G} = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹\frac{Nm^{2} }{kg^{2} }

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  F_{G} and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  F_{G} = \frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} } = 9.231 N

Therefore the acceleration due to gravity =  F_{G} /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²

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The answer is "143.74^{\circ} \ C , 8.36\ g, and \ 2.77\ \frac{K}{J}"

Explanation:

For point a:

Energy balance equation:

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From the above equation:

\frac{dU}{dt}=0-0+m_ih_i-0\\\\\Delta U=\int^{2}_{1}m_ih_idt\\\\

because the rate of air entering the tank that is h_i constant.

\Delta U = h_i \int^{2}_{1} m_i dt \\\\= h_i(m_2 -m_1)\\\\m_2u_2-m_1u_2=h_i(M_2-m_1)\\\\

Since the tank was initially empty and the inlet is constant hence, m_2u-0=h_1(m_2-0)\\\\m_2u_2=h_1m_2\\\\u_2=h_1\\\\

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T_1= 25^{\circ}\ C\ (298.15 \ K)\\\\\frac{h_{300 \ K}-h_{295\ K}}{300-295}= \frac{h_{300 \ K}-h_{1}}{300-295.15}

Substituting the value from ideal gas:

\frac{300.19-295.17}{300-295}=\frac{300.19-h_{i}}{300-298.15}\\\\h_i= 298.332 \ \frac{kJ}{kg}\\\\Now,\\\\h_i=u_2\\\\u_2=h_i=298.33\ \frac{kJ}{kg}

Follow the ideal gas table.

The u_2= 298.33\ \frac{kJ}{kg} and between temperature T =410 \ K \ and\  T=240\ K.

Interpolate

\frac{420-410}{u_{240\ k} -u_{410\ k}}=\frac{420-T_2}{u_{420 k}-u_2}

Substitute values from the table.

 \frac{420-410}{300.69-293.43}=\frac{420-T_2}{{u_{420 k}-u_2}}\\\\T_2=416.74\ K\\\\=143.74^{\circ} \ C\\\\

For point b:

Consider the ideal gas equation.  therefore, p is pressure, V is the volume, m is mass of gas. \bar{R} \ is\  \frac{R}{M} (M is the molar mass of the  gas that is 28.97 \ \frac{kg}{mol} and R is gas constant), and T is the temperature.

n=\frac{pV}{TR}\\\\

=\frac{(1.01 \times 10^5 \ Pa) \times (10\ L) (\frac{10^{-3} \ m^3}{1\ L})}{(416.74 K) (\frac{8.314 \frac{J}{mol.k} }{2897\ \frac{kg}{mol})}}\\\\=8.36\ g\\\\

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 Entropy is given by the following formula:

\Delta S = mC_v \In \frac{T_2}{T_1}\\\\=0.00836 \ kg \times 1.005 \times 10^{3} \In (\frac{416.74\ K}{298.15\ K})\\\\=2.77 \ \frac{J}{K}

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