Queston 6: Part A: At new moon, the Earth, Moon, and Sun are in a line, as indic

Question

Queston 6:

Part A: At new moon, the Earth, Moon, and Sun are in a line, as indicated in the figure(Figure 1) .

Find the magnitude of the net gravitational force exerted on the Earth.

Find the direction of the net gravitational force exerted on the Earth.

Find the magnitude of the net gravitational force exerted on the Moon

Find the direction of the net gravitational force exerted on the Moon.

Find the magnitude of the net gravitational force exerted on the Sun.

Find the direction of the net gravitational force exerted on the Sun.

Part B:

Gravity on Titan Titan is the largest moon of Saturn and the only moon in the solar system known to have a substantial atmosphere.

Find the acceleration due to gravity on Titan's surface, given that its mass is 1.35×1023kg and its radius is 2570 km.

Part C: Phobos, one of the moons of Mars, orbits at a distance of 9378 km from the center of the red planet.

What is the orbital period of Phobos?

Part D: The largest moon in the solar system is Ganymede, a moon of Jupiter. Ganymede orbits at a distance of 1.07×109m from the center of Jupiter with an orbital period of about 6.18×105s. Using this information, find the mass of Jupiter.

Explanation / Answer

partA

gravitational force is

FG = G*M*m/r2

magnitude of force between earth and moon

FG1 = 6.67*10-11 * 5.97*1024 *7.35*1022 / (3.84*108)2 = 1.98*1020 N

magnitude of force between earth and sun

FG2 =   6.67*10-11 * 5.97*1024 *1.99*1030 / (149.6*109)2 = 3.54*1022 N

magnitude of force between moon and sun

FG3 =   6.67*10-11 * 7.35*1022 *1.99*1030 / (149.6*109)2 = 4.38*1020 N

net force on earth

Fe = FG1 + FG2 = 1.98*1020 + 3.54*1022 = 3.56*1022 N

the direction of net gravitational force exerted on the earth is toward the moon and sun .hence direction is toward the sun.

net force on moon

Fm = FG3 - FG1 = 4.38*1020 - 1.98*1020 = 2.4*1020 N

the direction of net gravitational force exerted on the mon is toward the sun at it is greater than force by the earth toward it ,hence hence direction is toward the sun.

net force on sun

Fs = FG2 + FG3 = 4.38*1020 + 3.54*1022 = 3.58*1022 N

the direction of net gravitational force exerted on the sun is toward the moon and the earth .hence direction toward the earth moon system .

partB

ag = G*M/r2 = 6.67*1011 *1.35*1023 / (2570*103)2

ag = 1.19 m/s2

partC

The period of a satllite is related to its orbital radius by

T2 = 4*pi2*r3 /G*M

where G is the universal gravitational constant G = 6.67*10-11Nm2/kg2. We are given that Phobos hasan orbital radius of r = 9378 km = 9.378*106 m, and the mass of Mars is M = 6.42*1023 kg, so inserting these into our above expression for the period gives

T = 2*3.14*sqrt [ ( 9.378*106)3 / 6.67*1011 * 6.42*1023 ]

T =2.757*104 s

T = 2.757*104 /3600 = 7.660 h

partD

To find the gravitational pull use G*m*M / r^2 realize this gravitational froce is the centripetal force as the moon travels in a circle. centripetal froce is mv^2/r

therefore we have:
G*m*M/r^2 = m*v^2/r

now we use algebra to cancel m and r to simplify the equation:

GM/r = v^2

The only things we don't have are M and v. The period is the amount of time for one revolution. the length of the revolution is the circumfrence of the circle which we can get since we have the radius.

so one revolution= 1.07*10^9 *(2*pi) = 6.72 *10^9 m

the period is 6.18*10^5 s

using this information we get a velocity of 10874m/s

we plug this back into the original equation of G*M/r = v^2 to get

G*M/(1.07 *10^9) = (10874)^2

M= 1.9 * 10^27kg   

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