Question

A space probe is launched from Earth headed for deep space. At a distance of 10,000 miles from Earth's center, the gravitational force on it is 600lb. What is the size of the force when it is at each of the following distances from Earth's center? . A] 20,000 miles. B] 30,000 miles. C] 100,000 miles

Answer 1

 gravitational force varies based on 1/r^2
when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.
As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)
hope this helps

Answer 2

Answer: A)The size of the force when it is at 20,000 miles from earth center is 150 lb.

B) The size of the force when it is at 30,000 miles from earth center is 66.66 lb.

C) The size of the force when it is at 100,000 miles from earth center is 6 lb.

Explanation:

The Gravitational force is given as:

$F=\frac{G\times m_1\times m_2}{r^2}$

G= gravitational constant

$m_1,m_2$= masses of two objects exerting force on each other.

r = distance between the two objects

From the above expression it will be correct to write:

$F\propto \frac{1}{r^2}$

$F_1\times r^2_1=F_2\times r^2_2$

A) when the distance between the earth and space probe is 20,000 miles.

$F_1=600 lb,r_1=10,000 miles$

$F_2=?,r_2=20,000 miles$

$F_1\times r^2_1=F_2\times r^2_2$

$F_2=\frac{F_1\times r^2_1}{r^2_2}=\frac{600 lb\times 10,000\times 10,000}{20,000 miles\times 20,000 miles}=150 lb$

The size of the force when it is at 20,000 miles from earth center is 150 lb.

B) when the distance between the earth and space probe is 30,000 miles

$F_1=600 lb,r_1=10,000 miles$

$F_2=?,r_2=30,000 miles$

$F_1\times r^2_1=F_2\times r^2_2$

$F_2=\frac{F_1\times r^2_1}{r^2_2}=\frac{600 lb\times 10,000\times 10,000}{30,000 miles\times 30,000 miles}=66.66 lb$

The size of the force when it is at 30,000 miles from earth center is 66.66 lb.

C) when the distance between the earth and space probe is 100,000 miles

$F_1=600 lb,r_1=100,000 miles$

$F_2=?,r_2=100,000 miles$

$F_1\times r^2_1=F_2\times r^2_2$

$F_2=\frac{F_1\times r^2_1}{r^2_2}=\frac{600 lb\times 10,000\times 10,000}{100,000 miles\times 100,000 miles}=6 lb$

The size of the force when it is at 100,000 miles from earth center is 6 lb.