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e-lub [12.9K]
3 years ago
11

An object moves from point A to B to C to D and finally to A

Physics
1 answer:
Nataly [62]3 years ago
5 0

here's the solution,

The <em>radius</em> of the circle =<u> 3 km</u>

distance covered = <em>circumference</em> of the circle,

So, Circumference :

=》

2\pi r

=》

2 \times 3.14 \times 3

=》

18.84

(a). Distance covered by moving object is 18.84 km

(b). 0 km

now, Displacement of the object is 0 km, because displacement is the shortest distance from stating point to the destination, but the object returns back to the starting point, hence magnitude of displacement is 0.

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(a) 0.448

The gravitational potential energy of a satellite in orbit is given by:

U=-\frac{GMm}{r}

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G is the gravitational constant

M is the Earth's mass

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We can therefore write the ratio between the potentially energy of satellite B to that of satellite A as

\frac{U_B}{U_A}=\frac{-\frac{GMm}{R+h_B}}{-\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

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(b) 0.448

The kinetic energy of a satellite in orbit around the Earth is given by

K=\frac{1}{2}\frac{GMm}{r}

So, the ratio between the two kinetic energies is

\frac{K_B}{K_A}=\frac{\frac{1}{2}\frac{GMm}{R+h_B}}{\frac{1}{2}\frac{GMm}{R+h_A}}=\frac{R+h_A}{R+h_B}

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

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E_A=-\frac{1}{2}\frac{GMm}{R+h_A}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+5.97\cdot 10^6 m}=-4.65\cdot 10^8 J

For satellite B, we have

E_B=-\frac{1}{2}\frac{GMm}{R+h_B}=-\frac{1}{2}\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24}kg)(28.8 kg)}{6.37\cdot 10^6 m+21.2\cdot 10^6 m}=-2.08\cdot 10^8 J

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The difference between the energy of the two satellites is:

E_B-E_A=-2.08\cdot 10^8 J-(-4.65\cdot 10^8 J)=-2.57\cdot 10^8 J

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