Answer :
Certainly! Let's solve this step-by-step.
Given:
1. The mass of the celestial body is two times the mass of the Earth.
2. The radius of the celestial body is two times the radius of the Earth.
3. The acceleration due to gravity on Earth ([tex]$g_{\text{earth}}$[/tex]) is [tex]\(9.8\ \text{m/s}^2\)[/tex].
We are asked to find the acceleration due to gravity on the surface of the celestial body.
We use the formula for gravitational acceleration:
[tex]\[ g_{\text{new}} = G \frac{M_{\text{new}}}{R_{\text{new}}^2} \][/tex]
Here:
- [tex]\( G \)[/tex] is the gravitational constant.
- [tex]\( M_{\text{new}} \)[/tex] is the mass of the celestial body.
- [tex]\( R_{\text{new}} \)[/tex] is the radius of the celestial body.
Since the mass of the celestial body is two times the mass of the Earth ([tex]\( M_{\text{new}} = 2 M_{\text{earth}} \)[/tex]) and the radius is two times the radius of the Earth ([tex]\( R_{\text{new}} = 2 R_{\text{earth}} \)[/tex]), we substitute these values in the formula for gravitational acceleration:
[tex]\[ g_{\text{new}} = G \frac{2 M_{\text{earth}}}{(2 R_{\text{earth}})^2} \][/tex]
Simplify the expression inside the denominator:
[tex]\[ g_{\text{new}} = G \frac{2 M_{\text{earth}}}{4 R_{\text{earth}}^2} \][/tex]
Recognize that [tex]\( G \frac{M_{\text{earth}}}{R_{\text{earth}}^2} = g_{\text{earth}} \)[/tex].
Therefore:
[tex]\[ g_{\text{new}} = \frac{2 G M_{\text{earth}}}{4 R_{\text{earth}}^2} = \frac{2}{4} G \frac{M_{\text{earth}}}{R_{\text{earth}}^2} = \frac{1}{2} g_{\text{earth}} \][/tex]
Given that [tex]\( g_{\text{earth}} = 9.8\ \text{m/s}^2 \)[/tex]:
[tex]\[ g_{\text{new}} = \frac{1}{2} \times 9.8\ \text{m/s}^2 = 4.9\ \text{m/s}^2 \][/tex]
Thus, the acceleration due to gravity on the surface of the celestial body is [tex]\( 4.9\ \text{m/s}^2 \)[/tex].
Therefore, the correct answer is:
(ii) [tex]\( 4.9\ \text{m/s}^2 \)[/tex].
Given:
1. The mass of the celestial body is two times the mass of the Earth.
2. The radius of the celestial body is two times the radius of the Earth.
3. The acceleration due to gravity on Earth ([tex]$g_{\text{earth}}$[/tex]) is [tex]\(9.8\ \text{m/s}^2\)[/tex].
We are asked to find the acceleration due to gravity on the surface of the celestial body.
We use the formula for gravitational acceleration:
[tex]\[ g_{\text{new}} = G \frac{M_{\text{new}}}{R_{\text{new}}^2} \][/tex]
Here:
- [tex]\( G \)[/tex] is the gravitational constant.
- [tex]\( M_{\text{new}} \)[/tex] is the mass of the celestial body.
- [tex]\( R_{\text{new}} \)[/tex] is the radius of the celestial body.
Since the mass of the celestial body is two times the mass of the Earth ([tex]\( M_{\text{new}} = 2 M_{\text{earth}} \)[/tex]) and the radius is two times the radius of the Earth ([tex]\( R_{\text{new}} = 2 R_{\text{earth}} \)[/tex]), we substitute these values in the formula for gravitational acceleration:
[tex]\[ g_{\text{new}} = G \frac{2 M_{\text{earth}}}{(2 R_{\text{earth}})^2} \][/tex]
Simplify the expression inside the denominator:
[tex]\[ g_{\text{new}} = G \frac{2 M_{\text{earth}}}{4 R_{\text{earth}}^2} \][/tex]
Recognize that [tex]\( G \frac{M_{\text{earth}}}{R_{\text{earth}}^2} = g_{\text{earth}} \)[/tex].
Therefore:
[tex]\[ g_{\text{new}} = \frac{2 G M_{\text{earth}}}{4 R_{\text{earth}}^2} = \frac{2}{4} G \frac{M_{\text{earth}}}{R_{\text{earth}}^2} = \frac{1}{2} g_{\text{earth}} \][/tex]
Given that [tex]\( g_{\text{earth}} = 9.8\ \text{m/s}^2 \)[/tex]:
[tex]\[ g_{\text{new}} = \frac{1}{2} \times 9.8\ \text{m/s}^2 = 4.9\ \text{m/s}^2 \][/tex]
Thus, the acceleration due to gravity on the surface of the celestial body is [tex]\( 4.9\ \text{m/s}^2 \)[/tex].
Therefore, the correct answer is:
(ii) [tex]\( 4.9\ \text{m/s}^2 \)[/tex].
