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gizmo_the_mogwai [7]
3 years ago
14

Calculate the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun

. The particle is restricted to the line which joins the centers of the earth and the sun. Justify the two solutions physically. Note that: mearth = 5.976 x 1024 kg msun = 333,000 mearth Distance earth to sun = 149.6 Gm (giga-meters).
Physics
1 answer:
fredd [130]3 years ago
3 0

Answer:

149.34 Giga meter is the distance d from the center of the sun at which a particle experiences equal attractions from the earth and the sun.

Explanation:

Mass of earth = m = 5.976\times 10^{24} kg

Mass of Sun = M = 333,000 m

Distance between Earth and Sun = r = 149.6 gm =  1.496\times 10^{11} m[/tex]

1 giga meter = 10^{9} meter

Let the mass of the particle be m' which x distance from Sun.

Distance of the particle from Earth = (r-x)

Force between Sun and particle:

F=G\frac{M\times m'}{x^2}=G\frac{333,000 m\times m'}{x^2}

Force between Sun and particle:

F'=G\frac{mm'}{(r-x)^2}

Force on particle is equal:

F = F'

G\frac{333,000 m\times m'}{x^2}=G\frac{mm'}{(r-x)^2}

\frac{x}{r-x}=\sqrt{333,000} = ±577.06

Case 1:

\frac{x}{r-x}=577.06

x = 1.49\times 10^{11} m=149.34 Gm

Acceptable as the particle will lie in between the straight line joining Earth and Sun.

Case 2:

\frac{x}{r-x}=-577.06

x = 1.49\times 10^{11} m=149.86 Gm

Not acceptable as the particle will lie beyond on line extending straight from the Earth and Sun.

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