Physics, 11.01.2020 17:31, olguita82

Two earth satellites, a and b, each of mass m, are to be launched into circular orbits about earth's center. satellite a is to orbit at an altitude of 6380 km. satellite b is to orbit at an altitude of 22700 km. the radius of earth reis 6370 km. (a) what is the ratio of the potential energy of satellite b to that of satellite a, in orbit? (b) what is the ratio of the kinetic energy of satellite b to that of satellite a, in orbit? (c) which satellite (answer a or b) has the greater total energy if each has a mass of 35.0 kg? (d) by how much?

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answered: Graciouzgigi1394

(a) 0.439

The potential energy of a satellite in orbit is given by

$U=-\frac{GmM}{R+h}$

where

G is the gravitational constant

m is the mass of the satellite

M is the mass of the Earth

R is the Earth's radius

h is the altitude of the satellite

If we call

$U_A=-\frac{GmM}{R+h_A}$

the potential energy of satellite A, with

$h_A = 6380 km = 6.38\cdot 10^6 m$

being its altitude, and

$U_B=-\frac{GmM}{R+h_B}$

the potential energy of satellite B, with

$h_B = 22700 km = 22.7\cdot 10^6 m$

being the altitude of satellite B

and

$R=6370 km = 6.37 \cdot 10^6 m$ being the Earth's radius

The ratio between the potential energy of satellite B to that of satellite A will be

$\frac{U_B}{U_A}=\frac{R+h_A}{R+h_B}=\frac{6.37\cdot 10^6 m+6.38\cdot 10^6 m}{6.37\cdot 10^6 m+22.7\cdot 10^6 m}=0.439$

(b) 0.439

The kinetic energy of a satellite in orbit has a similar expression to the potential energy

$K=\frac{1}{2} \frac{GmM}{R+h}$

As before, if we call

$K_A=\frac{1}{2} \frac{GmM}{R+h_A}$

the kinetic energy of satellite A, with

$h_A = 6380 km = 6.38\cdot 10^6 m$

being its altitude, and

$K_B=\frac{1}{2} \frac{GmM}{R+h_B}$

the kinetic energy of satellite B, with

$h_B = 22700 km = 22.7\cdot 10^6 m$

being the altitude of satellite B,

the ratio between the kinetic energy of satellite B to that of satellite A is

$\frac{K_B}{K_A}=\frac{R+h_A}{R+h_B}=\frac{6.37\cdot 10^6 m+6.38\cdot 10^6 m}{6.37\cdot 10^6 m+22.7\cdot 10^6 m}=0.439$

The total energy of each satellite is given by the sum of the potential energy and the kinetic energy:

$E= U+K = -\frac{GMm}{R+h}+\frac{1}{2} \frac{GMm}{R+h}=-\frac{1}{2}\frac{GMm}{R+h}$

For satellite A we have:

$E_A = -\frac{1}{2}\frac{GMm}{R+h_A}$

While for satellite B we have

$E_B = -\frac{1}{2}\frac{GMm}{R+h_B}$

We see that the total energy is inversely proportional to the altitude of the satellite: therefore, the higher the satellite, the smaller the energy. So, satellite A will have the greater total energy (in magnitude), since h_A < h_B ; however, the value of the total energy is negative, so actually satellite B will have a greater energy than satellite A.

(d) $3.07\cdot 10^8 J$

The total energy of satellite A is

$E_A = -\frac{1}{2}\frac{GMm}{R+h_A}$

with

$h_A = 6380 km = 6.38\cdot 10^6 m$

while the total energy of satellite B is

$E_B = -\frac{1}{2}\frac{GMm}{R+h_B}$

with

$h_B = 22700 km = 22.7\cdot 10^6 m$

So the difference between the two energies is

$E_B - E_A = -\frac{1}{2}\frac{(6.67\cdot 10^{-11}(35 kg)(5.98\cdot 10^{24} kg)}{6.37\cdot 10^6 m +22.7\cdot 10^6 m}-(-\frac{1}{2}\frac{(6.67\cdot 10^{-11}(35 kg)(5.98\cdot 10^{24} kg)}{6.37\cdot 10^6 m +6.38\cdot 10^6 m})=3.07\cdot 10^8 J$

answered: Guest

This is what i found on google “what happens to the bond angle when you add or remove an electron domain? the bond angle decreases as you add domains 5. can you force the atoms into new configurations by pushing atoms around? note: we can think of a bond or a lone pair of electrons as a “domain” of electrons.” i hope this you

answered: Guest

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.

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Two earth satellites, a and b, each of mass m, are to be launched into circular orbits about earth's...