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Home » 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 , mass of the earth x . Neglect the effect of other planets etc. (orbital radius ).
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 \times 10^{30} \mathrm{~kg}, mass of the earth =6 x 10^{24} \mathrm{~kg}. Neglect the effect of other planets etc. (orbital radius =1.5 \times 10^{11} \mathrm{~m} ).
CBSE | Class 11 | Gravitation | NCERT | Physics
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 \times 10^{30} \mathrm{~kg}, mass of the earth =6 x 10^{24} \mathrm{~kg}. Neglect the effect of other planets etc. (orbital radius =1.5 \times 10^{11} \mathrm{~m} ).

Mass of Sun is given as M_{\text {sun }}=2 \times 10^{30} \mathrm{~kg}

Mass of Earth is given as M_{\text {earth }}=6 \times 10^{24} \mathrm{~kg}

Orbital radius is given as r=1.5 \times 10^{11} \mathrm{~m}

Let m be the mass of the rocket

Let P denote the location at which the gravitational pull on the rocket owing to Earth is greatest at a distance x.

The gravitational force exerted on the rocket at point P is equivalent using Newton’s law of gravitation.

\frac{G m M_{\text {sun }}}{(r-x)^{2}}=\frac{G m M_{\text {earth }}}{x^{2}}

\frac{(r-x)^{2}}{x^{2}}=\frac{M_{\text {sun }}}{M_{\text {carth }}}=\left(\frac{2 \times 10^{30}}{60 \times 10^{24}}\right)^{1 / 2}=577.35

1.5 \times 10^{11}-x=577.35 x

x=1.5 \times 10^{11} / 578.35=2.59 \times 10^{8} \mathrm{~m}

i

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