Two spacecraft in outer space attract each other with a force of 2N . What would the attractive force be if they were one-fourth as far apart?

2 Answers
Mar 13, 2018

#"32 N"#

Explanation:

Gravitational force of attraction between two objects of masses #"m"_1# and #"m"_2# separated by a distance (#"d"#) is given by

#"F" = ("Gm"_1"m"_2)/"d"^2#

When the speartion was #"d"#

#2 = ("Gm"_1"m"_2)/"d"^2#

When the separation was #"d"/4#

#"F" = ("Gm"_1"m"_2)/("d"/4)^2#

#"F" = 16 × ("Gm"_1"m"_2)/"d"^2#

#"F" = 16 × "2 N" = "32 N"#

Mar 13, 2018

The attractive force would be #32 N#.

Explanation:

The formula for the force of attraction between 2 bodies is

#F_g = (G*m_1*m_2)/r^2#

That is Newton's Universal Law of Gravitation. Let r be the original separation (when the attraction was 2 N), so #r/4# is the new separation.

Let's combine the product #G*m_1*m_2# into the constant k. #G and m_1 and m_2# will not change with this change in separation, so this is just for simplification #-# minimizing the confusion.

Then let's write the original and new attraction, #F_"g1" and F_"g2"# using the constant k.

#F_"g1" = k/r^2 = 2 N#

#F_"g2" = k/(r/4)^2 = k/(r^2/16)#

Now separate the expression that we found above for #F_"g1"#.

#F_"g2" = (16)*k/r^2 = (16)*F_"g1"#

So put in what #F_"g1"# is equal to

F_"g2" = 16*2 N = 32 N#

I hope this helps,
Steve