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A 525 kg satellite is in a circular orbit at an altitude of 575 km above the Earth's surface. Because of air friction, the satel

Dafna1 [17]

Answer:

$1.69\cdot 10^{10}J$

Explanation:

The total energy of the satellite when it is still in orbit is given by the formula

$E=-G\frac{mM}{2r}$

where

G is the gravitational constant

m = 525 kg is the mass of the satellite

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

r is the distance of the satellite from the Earth's center, so it is the sum of the Earth's radius and the altitude of the satellite:

$r=R+h=6370 km +575 km=6945 km=6.95\cdot 10^6 m$

So the initial total energy is

$E_i=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24} kg)}{2(6.95\cdot 10^6 m)}=-1.51\cdot 10^{10}J$

When the satellite hits the ground, it is now on Earth's surface, so

$r=R=6370 km=6.37\cdot 10^6 m$

so its gravitational potential energy is

$U = -G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(525 kg)(5.98\cdot 10^{24}kg)}{6.37\cdot 10^6 m}=-3.29\cdot 10^{10} J$

And since it hits the ground with speed

$v=1.90 km/s = 1900 m/s$

it also has kinetic energy:

$K=\frac{1}{2}mv^2=\frac{1}{2}(525 kg)(1900 m/s)^2=9.48\cdot 10^8 J$

So the total energy when the satellite hits the ground is

$E_f = U+K=-3.29\cdot 10^{10}J+9.48\cdot 10^8 J=-3.20\cdot 10^{10} J$

So the energy transformed into internal energy due to air friction is the difference between the total initial energy and the total final energy of the satellite:

$\Delta E=E_i-E_f=-1.51\cdot 10^{10} J-(-3.20\cdot 10^{10} J)=1.69\cdot 10^{10}J$

8 0

3 years ago

Synchronous communications satellites are placed in a circular orbit that is 3.59 107 m above the surface of the earth. What is

suter [353]

Answer: $0.223 m/s^{2}$

Explanation:

We can solve this with the Law of Universal Gravitation and knowing the acceleration due gravity $g$ of an object above the surface of the planet decreases with the distance (height) of this object from the center of the planet.