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help will give like -11 points V OSCOLPHYS2016 6.6.WA.052. My Notes Ask Your Teacher Determine the...

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-11 points V OSCOLPHYS2016 6.6.WA.052. My Notes Ask Your Teacher Determine the orbital speed, in m/s, of a satellite that cir
-11 points V OSCOLPHYS2016 6.6.WA.052. My Notes Ask Your Teacher Determine the orbital speed, in m/s, of a satellite that circles the Earth with a period of 2.10 x 104 s. The mass of the Earth is 5.97 x 1024 kg. m/s Additional Materials (a) What is the acceleration of gravity, in m/s, on the surface of the Moon? The mass of the moon is 7.35 x 1022 kg and its radius is 1.74 x 106m. 1.624 m/s2 (b) What is the acceleration of gravity, in m/s, on the surface (or outer limit) of Neptune? The mass of Neptune is 1.02 x 1026 kg and its radius is 2.48 x 10 m. co m/s² A heavier mass m, and a lighter mass m, are 17.5 cm apart and experience a gravitational force of attraction that is 8.70 x 10° N in magnitude. The two masses have a combined value of 5.60 kg. Determine the value of each individual mass. m , m2 = kg
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Answer #1

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Orbital Period is given by

T = 271 GME

(TPGM 1/3 =>r = 4+2 ) = (2.1 x 104)2(6.673 x 10-11) (5.97 x 1024) 1/3 472

r=1.6448442*107 m

Orbital speed is given by

v _ 2nr_20 X (1.6448442 x 10) (2.1 x 104) = 4921.4m/s

-------------------------------------------------------------------------------------------

a)

acceleration due to gravity is given by

g =GM/R2 = (6.673*10-11)(7.35*1022)/(1.74*106)2

g=1.62 m/s2

b)

For neptune

g=(6.673*10-11)(1.02*1026)/(2.48*107)2

g=11.07 m/s2

---------------------------------------------------------------------------------------------------------------

Given

m1+m2=5.6 Kg

m2=5.6-m1=

Gravitation force between two masses is given by

F=GM1M2/r2

8.7*10-9 = (6.673*10-11)m1(5.6-m1)/0.1752

5.6m1-m12 = 4

m12-5.6m1+4=0

m1=4.76 kg

m2=0.84 kg

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