Question #28033

On Earth's surface, we are about 6,500 km from Earth's center. The force of gravity on a certain person, at Earth's surface, causes his weight to be 160 lbs. How much will the person weigh at a distance of 13,000 km from Earth's center?

Expert's answer

g=G*M/R^2 - General formula to acceleration of gravity

G- gravity constant, R - distance, M Earth mass, in this problem G*M - const

P(weight of body)=m*g with this force body act to the Earth (weight isn't mass, it's depends from g)

P1=m*g1=160lbs Count weight of body for first case (from condition)

g1=const/(6,500*10^3)^2 cause g from first distance from the center

g2=const/(13,000*10^3)^2 from the second distance

note 6,500*10^3 as L (for convenience)

g1=const/L^2 &

g2=const/(2L)^2 rewrite equations

so g1=4*g2 (from two previous equations)

P2=mg2=(mg1)/4=40 lbs& (express g2 from g1 and substitute into equetion)

Answer weight of person at a distance of 13,000 km from Earth's center = 40 lbs

G- gravity constant, R - distance, M Earth mass, in this problem G*M - const

P(weight of body)=m*g with this force body act to the Earth (weight isn't mass, it's depends from g)

P1=m*g1=160lbs Count weight of body for first case (from condition)

g1=const/(6,500*10^3)^2 cause g from first distance from the center

g2=const/(13,000*10^3)^2 from the second distance

note 6,500*10^3 as L (for convenience)

g1=const/L^2 &

g2=const/(2L)^2 rewrite equations

so g1=4*g2 (from two previous equations)

P2=mg2=(mg1)/4=40 lbs& (express g2 from g1 and substitute into equetion)

Answer weight of person at a distance of 13,000 km from Earth's center = 40 lbs

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