# Answer to Question #17506 in Astronomy | Astrophysics for janice

Question #17506

two 1000kg asteroids are located 10m apart in the asteroid belt. waht is the gravitional attraction between them. B) if the mass of the first astroid triples, how would the force change?. C) if the mass of the first asteroid triples and then distance reduced by half, what would the new force be?

Expert's answer

two 1000kg asteroids are located 10m apart in the asteroid belt. What is the gravitational attraction between them.

F1 = G*M*M/r² =

= 6.67384×10^(-11)[m³/kg/s²] * 1000[kg] * 1000[kg] / (20[m])² =

& asymp; 1.6685*10^(-7) [N].

B) if the mass of the first astroid triples, how would the force change?

F2 = G*3*M*M/r² = 3F1.

So, the force will triple.

C) if the mass of the first asteroid triples and then distance reduced by half, what would the new force be?

F3 = G*3*M*M/(r/2)² =

= 6.67384×10^(-11)[m³/kg/s²] * 3 * 1000[kg] * 1000[kg] / (20[m]/2)² =

& asymp; 2.0022*10^(-6) [N].

F1 = G*M*M/r² =

= 6.67384×10^(-11)[m³/kg/s²] * 1000[kg] * 1000[kg] / (20[m])² =

& asymp; 1.6685*10^(-7) [N].

B) if the mass of the first astroid triples, how would the force change?

F2 = G*3*M*M/r² = 3F1.

So, the force will triple.

C) if the mass of the first asteroid triples and then distance reduced by half, what would the new force be?

F3 = G*3*M*M/(r/2)² =

= 6.67384×10^(-11)[m³/kg/s²] * 3 * 1000[kg] * 1000[kg] / (20[m]/2)² =

& asymp; 2.0022*10^(-6) [N].

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