Darian Hubbard
2022-07-22
Answered

What is the gravitational force between two identical 50,000 kg asteroids whose centers of mass are separated by 1000 m?

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Mireya Hoffman

Answered 2022-07-23
Author has **14** answers

Given data:

Two identical asteroids

${m}_{1}={m}_{2}=m=50,000\text{}kg$

Distance (r) = 1000 m

Required:

The gravitational force (${F}_{g}$)

We know that

Universal gravitational constant

$G=6.67\times {10}^{-11}\text{}N{m}^{2}/k{g}^{2}$

The gravitational force between two objects is given by

${F}_{g}=\frac{G{m}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{(6.67\times {10}^{-11}\text{}N{m}^{2}/k{g}^{2})(50000\text{}kg{)}^{2}}{(1000\text{}m{)}^{2}}\phantom{\rule{0ex}{0ex}}=1.6675\times {10}^{-7}\text{}N\phantom{\rule{0ex}{0ex}}\approx 1.67\times {10}^{-7}\text{}N$

The gravitational force between the asteroids is $1.67\times {10}^{-7}\text{}N$

Two identical asteroids

${m}_{1}={m}_{2}=m=50,000\text{}kg$

Distance (r) = 1000 m

Required:

The gravitational force (${F}_{g}$)

We know that

Universal gravitational constant

$G=6.67\times {10}^{-11}\text{}N{m}^{2}/k{g}^{2}$

The gravitational force between two objects is given by

${F}_{g}=\frac{G{m}^{2}}{{r}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{(6.67\times {10}^{-11}\text{}N{m}^{2}/k{g}^{2})(50000\text{}kg{)}^{2}}{(1000\text{}m{)}^{2}}\phantom{\rule{0ex}{0ex}}=1.6675\times {10}^{-7}\text{}N\phantom{\rule{0ex}{0ex}}\approx 1.67\times {10}^{-7}\text{}N$

The gravitational force between the asteroids is $1.67\times {10}^{-7}\text{}N$

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And when we're dealing with planets, we use a relation defined by the masses of two planets, distance squared and gravitational constant:

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