What is the size of the force of gravity between the Earth with a mass of 5.98 x #10^14# and the Sun that has a mass of 1.99 x #10^30# if the distance separating the earth and the sun is 1.50x#10^11# m ?

Question

Adrian Ayala · Accepted Answer

As per Law of Universal Gravitation the force of attraction #F_G# between two bodies of masses #M_1 and M_2# located at a distance distance #r# from each other is given as  #F_G =G (M_1.M_2)/r^2# .......(1) Where #G# is the proportionality constant.
It has the value #6.67408 xx 10^-11 m^3 kg^-1 s^-2# Inserting given values we get
#F_G =6.67408 xx 10^-11 xx(5.98 xx 10^24xx1.99 xx 10^30)/(1.50xx10^11)^2#
#=>F_G =3.53 xx 10^22N#
.-.-.-.-.-.-.-.-.- The mass of the earth appears to have been written incorrectly; the computations have fixed this.

Julia Aguilar · Answer

undefined The size of the force of gravity between the Earth and the Sun can be calculated using Newton's law of universal gravitation:

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

Where:
- $ F $ is the force of gravity
- $ G $ is the gravitational constant ($ 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 $)
- $ m_1 $ is the mass of the first object (mass of the Earth)
- $ m_2 $ is the mass of the second object (mass of the Sun)
- $ r $ is the distance between the centers of the two objects

Substituting the given values:

$$ F = \frac{{(6.674 \times 10^{-11}) \times (5.98 \times 10^{24}) \times (1.99 \times 10^{30})}}{{(1.50 \times 10^{11})^2}} $$

$$ F = \frac{{3.989 \times 10^{15}}}{{2.25 \times 10^{22}}} $$

$$ F \approx 1.78 \times 10^{22} \, \text{N} $$

Therefore, the size of the force of gravity between the Earth and the Sun is approximately $ 1.78 \times 10^{22} \, \text{N} $.