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.
The minimum energy required to change the orbit to another circular orbit is [tex]2.94 \times 10^8 \ J[/tex].
The given parameters;
The initial radius of the orbit is calculated as follows;
[tex]T^2 = \frac{4\pi^2 r^3}{GM} \\\\r = \sqrt[3]{\frac{T^2GM}{4\pi ^2} } \\\\r_1 = \sqrt[3]{\frac{7200^2 \times 6.67\times 10^{-11} \times 6 \times 10^{24}}{4\pi ^2} }\\\\r_1 = 8.07 \times 10^6 \ m[/tex]
The final radius of the orbit is calculated as follows;
[tex]r_2 = \sqrt[3]{\frac{10,800^2 \times 6.67\times 10^{-11} \times 6 \times 10^{24}}{4\pi ^2} }\\\\r_2 = 10.57 \times 10^6 \ m[/tex]
The minimum energy required to change the orbit to another circular orbit is calculated as follows;
[tex]E = \frac{1}{2} GMm(\frac{1}{r_1} - \frac{1}{r_2} )\\\\E = \frac{1}{2} \times 6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 50 (\frac{1}{8.07 \times 10^6 } - \frac{1}{10.57 \times 10^6} )\\\\E = 2.94 \times 10^8 \ J[/tex]
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