A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero? Mass of the sun = 2 x 1030 kg, mass of the earth = 6 x 1024 kg. Neglect the effect of other planets etc. (orbital radius = 1.5 x 1011 m).

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Now, let's tackle the problem at hand. When a rocket is fired from the Earth towards the Sun, we're essentially dealing with a scenario where gravitational forces from both celestial bodies come into play. The gravitational force between two masses can be calculated using Newton's law of universal gravitation:

F=G⋅m1⋅m2r2F=r2G⋅m1⋅m2

Where:

• FF is the gravitational force between the masses,
• GG is the gravitational constant (6.674×10−11 N m2/kg26.674×10−11N m2/kg2),
• m1m1 and m2m2 are the masses of the two objects,
• rr is the distance between the centers of the two masses.

At a certain distance from the Earth's center, the gravitational forces from the Earth and the Sun balance out such that the net gravitational force on the rocket becomes zero.

To find this point, we set up the gravitational force equations for the Earth and the Sun:

FEarth=FSunFEarth=FSun

G⋅MEarth⋅mr2=G⋅MSun⋅m(R−r)2r2G⋅MEarth⋅m=(R−r)2G⋅MSun⋅m

Where:

• MEarthMEarth and MSunMSun are the masses of the Earth and the Sun, respectively,
• mm is the mass of the rocket,
• RR is the distance from the Earth to the Sun (orbital radius).

We can simplify this equation and solve for rr:

MEarth(R−r)2=MSunr2(R−r)2MEarth=r2MSun

MEarth⋅r2=MSun⋅(R−r)2MEarth⋅r2=MSun⋅(R−r)2

Solving this equation will give us the distance from the Earth's center where the gravitational force on the rocket is zero. This calculation involves the masses of the Earth and the Sun, as well as the distance from the Earth to the Sun.

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