“2019年USAAAO决赛第8题”的版本间的差异
Jingsong Guo（讨论 | 贡献） （创建页面，内容为“==英文题目== 8. (15 points) In a rather weird universe, the gravitational constant G varies as a function of the scale factor aptq. G “ G0fpaq (5) Consider th…”） |
Jingsong Guo（讨论 | 贡献） |
||
| 第1行： | 第1行： | ||
==英文题目== | ==英文题目== | ||
| − | 8. (15 points) In a rather weird universe, the gravitational constant G varies as a function of the scale factor | + | 8. (15 points) In a rather weird universe, the gravitational constant G varies as a function of the scale factor $$a(0)$$. |
| + | |||
| + | $$G = G_0f(a)$$ | ||
| + | |||
| + | (5) Consider the model $$f(a)=e^{b(a-1)}$$ where $$b=2.09$$. | ||
| + | |||
| + | a) Assuming that the universe is flat, dark energy is absent, and the only constituent is matter, estimate the present age of this weird universe according to this model. Assume that the Friedmann equation: $$H(a)^2=H^2_0 (Ω_m+Ω_r+Ω_k+Ω_Λ)$$ still holds in this setting. | ||
| + | |||
| + | b) What is the behaviour of the age of the universe t as the scale factor $$a\rightarrow \infty$$ ? | ||
| + | |||
| + | Note that all parameters with subscript 0 indicate their present value. Take the value of Hubble’s constant as $$H0=67.8 kms^{-1}Mpc^{-1}$$ | ||
| + | |||
| + | Hint: You might need the following integrals: | ||
| + | |||
| + | $$\int_{0}^{\infty}x^2e^{-x^{2}}dx=\dfrac{\sqrt{\pi}}{4}$$ | ||
| + | |||
| + | $$\int_{0}^{1}x^2e^{-x^{2}}dx\approx 0.189471$$ | ||
2020年3月7日 (六) 18:08的最新版本
英文题目
8. (15 points) In a rather weird universe, the gravitational constant G varies as a function of the scale factor $$a(0)$$.
$$G = G_0f(a)$$
(5) Consider the model $$f(a)=e^{b(a-1)}$$ where $$b=2.09$$.
a) Assuming that the universe is flat, dark energy is absent, and the only constituent is matter, estimate the present age of this weird universe according to this model. Assume that the Friedmann equation: $$H(a)^2=H^2_0 (Ω_m+Ω_r+Ω_k+Ω_Λ)$$ still holds in this setting.
b) What is the behaviour of the age of the universe t as the scale factor $$a\rightarrow \infty$$ ?
Note that all parameters with subscript 0 indicate their present value. Take the value of Hubble’s constant as $$H0=67.8 kms^{-1}Mpc^{-1}$$
Hint: You might need the following integrals:
$$\int_{0}^{\infty}x^2e^{-x^{2}}dx=\dfrac{\sqrt{\pi}}{4}$$
$$\int_{0}^{1}x^2e^{-x^{2}}dx\approx 0.189471$$
