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3 years ago

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A satellite orbits a planet of unknown mass in a circular orbit of radius 2.3 x 104 km. The gravitational force on the satellite

from the planet is 6600 N. What is the kinetic energy of the satellite

1 answer:

sladkih [1.3K]3 years ago

7 0

Answer:

The kinetic energy is $KE = 7.59 *10^{10} \ J$

Explanation:

From the question we are told that

The radius of the orbit is $r = 2.3 *10^{4} \ km = 2.3 *10^{7} \ m$

The gravitational force is $F_g = 6600 \ N$

The kinetic energy of the satellite is mathematically represented as

$KE = \frac{1}{2} * mv^2$

where v is the speed of the satellite which is mathematically represented as

$v = \sqrt{\frac{G M}{r^2} }$

=> $v^2 = \frac{GM }{r}$

substituting this into the equation

$KE = \frac{ 1}{2} *\frac{GMm}{r}$

Now the gravitational force of the planet is mathematically represented as

$F_g = \frac{GMm}{r^2}$

Where M is the mass of the planet and m is the mass of the satellite

Now looking at the formula for KE we see that we can represent it as

$KE = \frac{ 1}{2} *[\frac{GMm}{r^2}] * r$

=> $KE = \frac{ 1}{2} *F_g * r$

substituting values

$KE = \frac{ 1}{2} *6600 * 2.3*10^{7}$

$KE = 7.59 *10^{10} \ J$

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The moment of inertial is mathematically represented as

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$\alpha = \frac{torque }{moment \ of \ inertia} = \frac{F_d * R_d}{I}$

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Substituting values

$\alpha _t = (0.200) (7.5)$

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The magnitude of resultant acceleration at that point is

$a = \sqrt{\alpha_r ^2+ \alpha_t^2 }$

Substituting values

$a = \sqrt{(3.7699)^2 + (1.5)^2}$

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