A planet with a mass of (7*10^21) kg is in a circular orbit around a star with a mass of (2*10^30) kg. The planet has an orbital radius of (3*10^10) m.

(a) What is the linear orbital velocity of the planet?
(b) What is the period of the planet's orbit?
(c) What is the total mechanical energy of the planet?

To calculate the answers to these questions, we can use various formulas related to circular motion and gravitational force. Let's go step by step:

(a) The linear orbital velocity of the planet can be determined using the formula for centripetal acceleration:

v = sqrt(G * M / r)

Where:
v - linear orbital velocity
G - gravitational constant (approximately 6.674 * 10^-11 m^3/kg/s^2)
M - mass of the star
r - orbital radius of the planet

Plugging in the given values:
M = 2 * 10^30 kg
r = 3 * 10^10 m

Calculating:
v = sqrt((6.674 * 10^-11 m^3/kg/s^2) * (2 * 10^30 kg) / (3 * 10^10 m))

Simplifying:
v = sqrt((13.348 * 10^19 m^4/kg/s^2) * (2 * 10^30 kg) / (3 * 10^10 m))

Further simplifying:
v = sqrt((26.696 * 10^49 m^5/kg/s^2) / (3 * 10^10 m))

And finally:
v = sqrt(8.89867 * 10^38 m^4/s^2)

Therefore, the linear orbital velocity of the planet is approximately 2.98216 * 10^19 m/s.

(b) The period of the planet's orbit can be calculated using the formula:

T = (2πr) / v

Where:
T - period of the orbit (time taken for one complete revolution)
r - orbital radius of the planet
v - linear orbital velocity of the planet

Plugging in the given values:
r = 3 * 10^10 m
v = 2.98216 * 10^19 m/s

Calculating:
T = (2π * (3 * 10^10 m)) / (2.98216 * 10^19 m/s)

Simplifying:
T = (6π * 10^40 m) / (2.98216 * 10^19 m/s)

Further simplifying:
T = (18.84956 * 10^40 m) / (2.98216 * 10^19 m/s)

And finally:
T = 6.32468 * 10^21 s

Therefore, the period of the planet's orbit is approximately 6.32468 * 10^21 seconds.

(c) The total mechanical energy of the planet in its orbit can be calculated using the formula:

E = - (G * M * m) / (2r)

Where:
E - total mechanical energy
G - gravitational constant
M - mass of the star
m - mass of the planet
r - orbital radius of the planet

Plugging in the given values:
G = 6.674 * 10^-11 m^3/kg/s^2
M = 2 * 10^30 kg
m = 7 * 10^21 kg
r = 3 * 10^10 m

Calculating:
E = - (6.674 * 10^-11 m^3/kg/s^2) * (2 * 10^30 kg) * (7 * 10^21 kg) / (2 * (3 * 10^10 m))

Simplifying:
E = - (13.348 * 10^19 m^4/kg/s^2) * (7 * 10^21 kg) / (2 * (3 * 10^10 m))

Further simplifying:
E = - (93.436 * 10^40 m^5/kg/s^2) / (6 * 10^10 m)

And finally:
E = -15.5727 * 10^29 J

Therefore, the total mechanical energy of the planet in its orbit is approximately -1.55727 * 10^30 Joules. Note that the negative sign indicates that the energy is in a bound (stable) orbit.