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Flura [38]
3 years ago
14

A 50-kg satellite circles the Earth in an orbit with a period of 120 min. What minimum energy is required to change the orbit to

another circular orbit with a period of 180 min? (Earth: radius = 6.4 106 m, mass = 6.0 1024 kg)
Physics
1 answer:
uysha [10]3 years ago
6 0

Answer: 2.94×10^8 J

Explanation:

Using the relation

T^2 = (4π^2/GMe) r^3

Where v= velocity

r = radius

T = period

Me = mass of earth= 6×10^24

G = gravitational constant= 6.67×10^-11

4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]

= 0.9865 x 10^-13

Therefore,

T^2 = (0.9865 × 10^-13) × r^3

r^3 = 1/(0.9865 × 10^-13) ×T^2

r^3 = (1.014 x 10^13) × T^2

To find r1 and r2

T1 = 120min = 120*60 = 7200s

T2 = 180min = 180*60= 10800s

Therefore,

r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m

r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m

Required Mechanical energy

= - GMem/2 [1/r2 - 1/r1]

= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]

= (2001 x 10^7)/2 * (0.1239 - 0.0945)

= (1000.5 × 10^7) × 0.0294

= 29.4147 × 10^7 J

= 2.94 x 10^8 J.

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KatRina [158]

Answer:

a)  0.28 m or 28 cm is the minimum  height above ground the fish reaches.

b)  at the height of 0.484 m height , the pufferfish will eventually come to rest.

c) There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = 23.72J

Spring potential energy   = 0.384 J

Explanation:

Given that :

Mass of the pufferfish m =5kg

initial height of the fish h =5.5m

length of the spring l =0.5m

Spring constant K =3000N/m

a)

Assuming no energy loss to friction, what is the minimum height above the ground that the pufferfish reaches?

Lets assume that the minimum height the fish reaches is = x meters

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Initial total energy = final total energy

Gravitational potential energy =

Gravitational potential energy' + Spring potential energy (kinetic energy is zero in both cases)

mgh = mgx + \frac{1}{2}K(l-x)^2

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(5)(9.8)(5.5) = (5)(9.5)(x) + \frac{1}{2}(3000)(0.5-x)^2

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269.5 - 375 = 47.5 x - 1500 x²

-105.5 = 47.5 x - 1500 x²

-105.5 + 1500 x² - 47.5 x = 0

1500 x² - 47.5 x - 105.5 = 0

By using quadratic equation and taking the positive value;

x = 0.28 m or 28 cm is the minimum height above ground the fish reaches.

b)

At the equilibrium position the weight of fish will be equal to the force applied by the spring thus

mg = kx

substituting  our given values ; we have:

(5)(9.8) = 3000x

x = 61.22

x = 0.016m  : so this is the compression in the spring

Now; to determine the height  the pufferfish gets to before  it eventually come to rest; we have

(0.5-0.016) m = 0.484m

therefore, at the height of 0.484 m height , the pufferfish will eventually come to rest.

c)

There exists  two types of energy remain at the equilibrium point in the system. These are :

Gravitational potential energy  = mgh' = (5)(9.8)(0.484)

= 23.72J

and spring potential energy  

=\frac{1}{2}Kx^2\\ = \frac{1}{2}(3000)(0.016)^2\\= 0.384J

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<h3><u>Answer;</u></h3>

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