Answer :
We start by noting that for an object in a circular orbit, the gravitational force acting as the centripetal force is given by
[tex]$$
\frac{GM_{\text{sun}}m}{r^2} = \frac{mv^2}{r},
$$[/tex]
where
- [tex]$G$[/tex] is the gravitational constant,
- [tex]$M_{\text{sun}}$[/tex] is the mass of the Sun,
- [tex]$m$[/tex] is the mass of the planet,
- [tex]$v$[/tex] is the orbital speed, and
- [tex]$r$[/tex] is the orbital radius.
Notice that the planet’s mass [tex]$m$[/tex] cancels out.
Solving for [tex]$r$[/tex], we get
[tex]$$
r = \frac{GM_{\text{sun}}}{v^2}.
$$[/tex]
The known quantities are:
- [tex]$G = 6.67430 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}$[/tex],
- [tex]$M_{\text{sun}} = 1.9885 \times 10^{30} \, \text{kg}$[/tex],
- [tex]$v = 19.77 \, \text{km/s}$[/tex], which we convert to meters per second:
[tex]$$
v = 19.77 \times 1000 = 19770 \, \text{m/s}.
$$[/tex]
Next, we compute the product [tex]$GM_{\text{sun}}$[/tex]:
[tex]$$
GM_{\text{sun}} = \left(6.67430 \times 10^{-11}\right) \times \left(1.9885 \times 10^{30}\right) \approx 1.3272 \times 10^{20} \, \text{m}^3\text{s}^{-2}.
$$[/tex]
Now, compute the square of the speed:
[tex]$$
v^2 = \left(19770 \, \text{m/s}\right)^2 \approx 3.90853 \times 10^{8} \, \text{m}^2\text{s}^{-2}.
$$[/tex]
Substitute these values into the expression for [tex]$r$[/tex]:
[tex]$$
r = \frac{1.3272 \times 10^{20}}{3.90853 \times 10^{8}} \approx 3.40 \times 10^{11} \, \text{m}.
$$[/tex]
Since the computed distance is approximately [tex]$3.40 \times 10^{11}$[/tex] m, this is closest to option (D):
[tex]$$
\boxed{3.53 \times 10^{11} \, \text{m}}.
$$[/tex]
Thus, the correct answer is option (D).
[tex]$$
\frac{GM_{\text{sun}}m}{r^2} = \frac{mv^2}{r},
$$[/tex]
where
- [tex]$G$[/tex] is the gravitational constant,
- [tex]$M_{\text{sun}}$[/tex] is the mass of the Sun,
- [tex]$m$[/tex] is the mass of the planet,
- [tex]$v$[/tex] is the orbital speed, and
- [tex]$r$[/tex] is the orbital radius.
Notice that the planet’s mass [tex]$m$[/tex] cancels out.
Solving for [tex]$r$[/tex], we get
[tex]$$
r = \frac{GM_{\text{sun}}}{v^2}.
$$[/tex]
The known quantities are:
- [tex]$G = 6.67430 \times 10^{-11} \, \text{m}^3\text{kg}^{-1}\text{s}^{-2}$[/tex],
- [tex]$M_{\text{sun}} = 1.9885 \times 10^{30} \, \text{kg}$[/tex],
- [tex]$v = 19.77 \, \text{km/s}$[/tex], which we convert to meters per second:
[tex]$$
v = 19.77 \times 1000 = 19770 \, \text{m/s}.
$$[/tex]
Next, we compute the product [tex]$GM_{\text{sun}}$[/tex]:
[tex]$$
GM_{\text{sun}} = \left(6.67430 \times 10^{-11}\right) \times \left(1.9885 \times 10^{30}\right) \approx 1.3272 \times 10^{20} \, \text{m}^3\text{s}^{-2}.
$$[/tex]
Now, compute the square of the speed:
[tex]$$
v^2 = \left(19770 \, \text{m/s}\right)^2 \approx 3.90853 \times 10^{8} \, \text{m}^2\text{s}^{-2}.
$$[/tex]
Substitute these values into the expression for [tex]$r$[/tex]:
[tex]$$
r = \frac{1.3272 \times 10^{20}}{3.90853 \times 10^{8}} \approx 3.40 \times 10^{11} \, \text{m}.
$$[/tex]
Since the computed distance is approximately [tex]$3.40 \times 10^{11}$[/tex] m, this is closest to option (D):
[tex]$$
\boxed{3.53 \times 10^{11} \, \text{m}}.
$$[/tex]
Thus, the correct answer is option (D).
