The period of the orbit would increase as well
Explanation:
We can answer this question by applying Kepler's third law, which states that:
"The square of the orbital period of a planet around the Sun is proportional to the cube of the semi-major axis of its orbit"
Mathematically,
Where
T is the orbital period
a is the semi-major axis of the orbit
In this problem, the question asks what happens if the distance of the Earth from the Sun increases. Increasing this distance means increasing the semi-major axis of the orbit, 
: but as we saw from the previous equation, the orbital period of the Earth is proportional to , therefore as increases, T increases as well.
Therefore, the period of the orbit would increase.
Learn more about Kepler's third law:
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Let the rise in temperature be 
The expansion in length due to change in temperature is given by the expression lαΔt , where l is the length, α is the coefficient of linear expansion, Δt is the change in temperature.
Here l = 93 m, α = 
, and Δt =
So expansion in length = 93**5 = 0.007905 m = 
So order of magnitude in change in length = -3
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Answer:</h2><h2>
The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/
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Explanation:
A meteoroid is in a circular orbit 600 km above the surface of a distant planet.
Mass of the planet = mass of earth = 5.972 x 
Kg
Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km
The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?
By formula, g = 
where g is the acceleration due to the gravity
G is the universal gravitational constant = 6.67 x 
M is the mass of the planet
r is the radius of the planet
Substituting the values, we get
g = 
g = 12.12 m/
The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/
Average speed = total distance travelled ÷ total time taken
AS = (75km + 68km) ÷ (1hr + 2hr)
As = 143km ÷ 3hr
AS = 47.66667 km/hr
AS = 47.7 km/hr (3sf)
