Question

Determine the force of gravitational attraction between the earth (m = 5.98 x 1024 kg) and a 70-kg physics student if the student is in an airplane at 40000 feet above earth's surface. This would place the student a distance of 6.39 x 106 m from earth's center.

Answer 1

Answer:

$F=683.790939$N

Explanation:

From the Question we are told that

Earth mass $m_1= 5.98 * 10^2^4 kg$

Students mass $m_2=70kg$

Distance of student from earth center Radius $r=6.39 * 10^6 m$

Generally the equation for Force of attraction b/w the earth and the boy is mathematically given by

 $F=\frac{Gm_1m_2}{r^2}$

 $F=\frac{6.67.10^-^1^1*( 5.98 x 10^2^4)(70)}{(6.39 *10^6)^2}$

Therefore the force of attraction b/w the student and the earth is

 $F=683.790939N$

 $F\approx 684N$